Content check — Calculus Vol. 1, Chapter 2: Limits

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 2: Limits

146 problems · 37 figures

Checkpoint 2.1 §2.1 source ↗

Estimate the slope of the tangent line (rate of change) to

f (x) = x^{2}

x = 1

by finding the slope of the secant line through

(1, 1)

and the point

(\frac{5}{4}, \frac{25}{16})

on the graph of

f (x) = x^{2} .

Answer

2.25

Checkpoint 2.2 §2.1 source ↗

An object moves along a coordinate axis so that its position at time t is given by

s (t) = t^{3} .

Estimate its instantaneous velocity at time

t = 2

by computing its average velocity over the time interval

[2, 2.001] .

Answer

12.006001

Checkpoint 2.3 §2.1 source ↗

Estimate the area between the x-axis and the graph of $f (x) = x^{2} + 1$ over the interval $[0, 3]$ by using the three rectangles shown here:

A graph of the same parabola f(x) = x^2 + 1, but with a different shading strategy over the interval [0,3]. This time, the shaded rectangles are given the height of the taller corner that could intersect with the function. As such, the rectangles go higher than the height of the function.

Answer

17 unit²

2-1 §2.1 source ↗

For the following exercises, points $P (1, 2)$ and $Q (x, y)$ are on the graph of the function $f (x) = x^{2} + 1 .$

[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $Q (x, y),$ and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.

x	y	$Q (x, y)$	m_sec
1.1	a.	e.	i.
1.01	b.	f.	j.
1.001	c.	g.	k.
1.0001	d.	h.	l.

Answer

a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

2-3 §2.1 source ↗

For the following exercises, points $P (1, 2)$ and $Q (x, y)$ are on the graph of the function $f (x) = x^{2} + 1 .$

Use the value in the preceding exercise to find an equation of the tangent line at point P. Graph $f (x)$ and the tangent line.

Answer

y = 2 x

2-5 §2.1 source ↗

For the following exercises, points $P (1, 1)$ and $Q (x, y)$ are on the graph of the function $f (x) = x^{3} .$

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $x = 1 .$

Answer

2-7 §2.1 source ↗

For the following exercises, points $P (4, 2)$ and $Q (x, y)$ are on the graph of the function $f (x) = \sqrt{x} .$

x	y	$Q (x, y)$	m_sec
4.1	a.	e.	i.
4.01	b.	f.	j.
4.001	c.	g.	k.
4.0001	d.	h.	l.

Answer

a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

2-9 §2.1 source ↗

For the following exercises, points $P (4, 2)$ and $Q (x, y)$ are on the graph of the function $f (x) = \sqrt{x} .$

Use the value in the preceding exercise to find an equation of the tangent line at point P.

Answer

y = \frac{x}{4} + 1

2-11 §2.1 source ↗

For the following exercises, points $P (1.5, 0)$ and $Q (φ, y)$ are on the graph of the function $f (φ) =$ $cos (π φ) .$

Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $φ = 1.5 .$

Answer

2-13 §2.1 source ↗

For the following exercises, points $P (−1, −1)$ and $Q (x, y)$ are on the graph of the function $f (x) = \frac{1}{x} .$

x	y	$Q (x, y)$	m_sec
−1.05	a.	e.	i.
−1.01	b.	f.	j.
−1.005	c.	g.	k.
−1.001	d.	h.	l.

Answer

a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100

2-15 §2.1 source ↗

For the following exercises, points $P (−1, −1)$ and $Q (x, y)$ are on the graph of the function $f (x) = \frac{1}{x} .$

Use the value in the preceding exercise to find an equation of the tangent line at point P.

Answer

y = − x - 2

2-17 §2.1 source ↗

For the following exercises, the position function of a ball dropped from the top of a 200-meter tall building is given by $s (t) =$ $200 -$ $4.9 t^{2},$ where position s is measured in meters and time t is measured in seconds. Round your answer to eight significant digits.

Use the preceding exercise to guess the instantaneous velocity of the ball at $t = 5$ sec.

Answer

−49 m/sec (velocity of the ball is 49 m/sec downward)

2-19 §2.1 source ↗

For the following exercises, consider a stone tossed into the air from ground level with an initial velocity of 15 m/sec. Its height in meters at time t seconds is $h (t) =$ $15 t -$ $4.9 t^{2} .$

Use the preceding exercise to guess the instantaneous velocity of the stone at $t = 1$ sec.

Answer

5.2 m/sec

2-21 §2.1 source ↗

For the following exercises, consider a rocket shot into the air that then returns to Earth. The height of the rocket in meters is given by $h (t) =$ $600 +$ $78.4 t -$ $4.9 t^{2},$ where t is measured in seconds.

Use the preceding exercise to guess the instantaneous velocity of the rocket at $t = 9$ sec.

Answer

−9.8 m/sec

2-23 §2.1 source ↗

For the following exercises, consider an athlete running a 40-m dash. The position of the athlete is given by $d (t) =$ $\frac{t^{3}}{6} +$ $4 t,$ where d is the position in meters and t is the time elapsed, measured in seconds.

Use the preceding exercise to guess the instantaneous velocity of the runner at $t = 2$ sec.

Answer

6 m/sec

2-25 §2.1 source ↗

For the following exercises, consider the function $f (x) = | x | .$

Use the preceding exercise to find the aproximate value of the area between the x-axis and the graph of f over the interval $[−1, 2]$ using rectangles. For the rectangles, use the square units, and approximate both above and below the lines. Use geometry to find the exact answer.

Answer

Under, 1 unit²; over: 4 unit². The exact area of the two triangles is

\frac{1}{2} (1) (1) +

\frac{1}{2} (2) (2) =

2.5 {units}^{2} .

2-27 §2.1 source ↗

For the following exercises, consider the function $f (x) = \sqrt{1 - x^{2}} .$ (Hint: This is the upper half of a circle of radius 1 positioned at $(0, 0) .)$

Use the preceding exercise to find the aproximate area between the x-axis and the graph of f over the interval $[−1, 1]$ using rectangles. For the rectangles, use squares 0.4 by 0.4 units, and approximate both above and below the lines. Use geometry to find the exact answer.

Answer

Under, 0.96 unit²; over, 1.92 unit². The exact area of the semicircle with radius 1 is

\frac{π {(1)}^{2}}{2} = \frac{π}{2}

unit².

2-29 §2.1 source ↗

For the following exercises, consider the function $f (x) = − x^{2} + 1 .$

Approximate the area of the region between the x-axis and the graph of f over the interval $[−1, 1] .$

Answer

Approximately 1.3333333 unit²

Checkpoint 2.4 §2.2 source ↗

Estimate

lim_{x \to 1} \frac{\frac{1}{x} - 1}{x - 1}

using a table of functional values. Use a graph to confirm your estimate.

Answer

lim_{x \to 1} \frac{\frac{1}{x} - 1}{x - 1} =

−1

Checkpoint 2.5 §2.2 source ↗

Use the graph of

h (x)

in Figure 2.16 to evaluate

lim_{x \to 2} h (x),

if possible.

A graph of the function h(x), which is a parabola graphed over [-2.5, 5]. There is an open circle where the vertex should be at the point (2,-1). — Figure 2.16 in the book ↗

Answer

lim_{x \to 2} h (x) =

−1 .

Checkpoint 2.6 §2.2 source ↗

Use a table of functional values to evaluate

lim_{x \to 2} \frac{| x^{2} - 4 |}{x - 2},

if possible.

Answer

lim_{x \to 2} \frac{| x^{2} - 4 |}{x - 2}

does not exist.

Checkpoint 2.7 §2.2 source ↗

Use a table of functional values to estimate the following limits, if possible.

a. $lim_{x \to 2^{-}} \frac{| x^{2} - 4 |}{x - 2}$

b. $lim_{x \to 2^{+}} \frac{| x^{2} - 4 |}{x - 2}$

Answer

lim_{x \to 2^{-}} \frac{| x^{2} - 4 |}{x - 2} =

−4;

lim_{x \to 2^{+}} \frac{| x^{2} - 4 |}{x - 2} =

4

Checkpoint 2.8 §2.2 source ↗

Evaluate each of the following limits, if possible. Use a table of functional values and graph $f (x) = 1 / x^{2}$ to confirm your conclusion.

a. $lim_{x \to 0^{-}} \frac{1}{x^{2}}$

b. $lim_{x \to 0^{+}} \frac{1}{x^{2}}$

c. $lim_{x \to 0} \frac{1}{x^{2}}$

Answer

lim_{x \to 0^{-}} \frac{1}{x^{2}} =

+ \infty;

lim_{x \to 0^{+}} \frac{1}{x^{2}} =

+ \infty;

lim_{x \to 0} \frac{1}{x^{2}} =

+ \infty

Checkpoint 2.9 §2.2 source ↗

Evaluate each of the following limits. Identify any vertical asymptotes of the function $f (x) =$ $\frac{1}{{(x - 2)}^{3}} .$

a. $lim_{x \to 2^{-}} \frac{1}{{(x - 2)}^{3}}$

b. $lim_{x \to 2^{+}} \frac{1}{{(x - 2)}^{3}}$

c. $lim_{x \to 2} \frac{1}{{(x - 2)}^{3}}$

Answer

lim_{x \to 2^{-}} \frac{1}{{(x - 2)}^{3}} =

− \infty;

lim_{x \to 2^{+}} \frac{1}{{(x - 2)}^{3}} =

+ \infty;

lim_{x \to 2} \frac{1}{{(x - 2)}^{3}}

DNE. The line

x = 2

is the vertical asymptote of

f (x) =

1 / {(x - 2)}^{3} .

Checkpoint 2.10 §2.2 source ↗

Evaluate $lim_{x \to 1} f (x)$ for $f (x)$ shown here:

A graph of a piecewise function. The first segment curves from the third quadrant to the first, crossing through the second quadrant. Where the endpoint would be in the first quadrant is an open circle. The second segment starts at a closed circle a few units below the open circle. It curves down from quadrant one to quadrant four.

Answer

Does not exist.

2-31 §2.2 source ↗

For the following exercises, consider the function $f (x) =$ $\frac{x^{2} - 1}{| x - 1 |} .$

What do your results in the preceding exercise indicate about the two-sided limit $lim_{x \to 1} f (x) ?$ Explain your response.

Answer

lim_{x \to 1} f (x)

does not exist because

lim_{x \to 1^{-}} f (x) =

−2 \neq lim_{x \to 1^{+}} f (x) =

2 .

2-33 §2.2 source ↗

For the following exercises, consider the function $f (x) =$ ${(1 + x)}^{1 / x} .$

What does the table of values in the preceding exercise indicate about the function $f (x) =$ ${(1 + x)}^{1 / x} ?$

Answer

lim_{x \to 0} {(1 + x)}^{1 / x} =

2.7183

2-35 §2.2 source ↗

In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.

[T] $lim_{x \to 0} \frac{sin 2 x}{x};$ $±0.1,$ $±0.01,$ $±0.001,$ $±.0001$

x	$\frac{sin 2 x}{x}$	x	$\frac{sin 2 x}{x}$
−0.1	a.	0.1	e.
−0.01	b.	0.01	f.
−0.001	c.	0.001	g.
−0.0001	d.	0.0001	h.

Answer

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999;

lim_{x \to 0} \frac{sin 2 x}{x} =

2

2-37 §2.2 source ↗

In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.

Use the preceding two exercises to conjecture (guess) the value of the following limit: $lim_{x \to 0} \frac{sin a x}{x}$ for a, a positive real value.

Answer

lim_{x \to 0} \frac{sin a x}{x} =

a

2-39 §2.2 source ↗

[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

$lim_{x \to 1} (1 - 2 x)$

x	$1 - 2 x$	x	$1 - 2 x$
0.9	a.	1.1	e.
0.99	b.	1.01	f.
0.999	c.	1.001	g.
0.9999	d.	1.0001	h.

Answer

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000;

lim_{x \to 1} (1 - 2 x) =

−1

2-41 §2.2 source ↗

[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

$lim_{z \to 0} \frac{z - 1}{z^{2} (z + 3)}$

z	$\frac{z - 1}{z^{2} (z + 3)}$	z	$\frac{z - 1}{z^{2} (z + 3)}$
−0.1	a.	0.1	e.
−0.01	b.	0.01	f.
−0.001	c.	0.001	g.
−0.0001	d.	0.0001	h.

Answer

a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889

lim_{x \to 0} \frac{z - 1}{z^{2} (z + 3)} =

− \infty

2-43 §2.2 source ↗

[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

$lim_{x \to 2} \frac{1 - \frac{2}{x}}{x^{2} - 4}$

x	$\frac{1 - \frac{2}{x}}{x^{2} - 4}$	x	$\frac{1 - \frac{2}{x}}{x^{2} - 4}$
1.9	a.	2.1	e.
1.99	b.	2.01	f.
1.999	c.	2.001	g.
1.9999	d.	2.0001	h.

Answer

a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063;

∴ lim_{x \to 2} \frac{1 - \frac{2}{x}}{x^{2} - 4} =

0.1250 =

\frac{1}{8}

2-45 §2.2 source ↗

[T] In the following exercises, set up a table of values and round to eight significant digits. Based on the table of values, make a guess about what the limit is. Then, use a calculator to graph the function and determine the limit. Was the conjecture correct? If not, why does the method of tables fail?

$lim_{α \to 0^{+}} \frac{1}{α} cos (\frac{π}{α})$

$α$	$\frac{1}{α} cos (\frac{π}{α})$
0.1	a.
0.01	b.
0.001	c.
0.0001	d.

Answer

a. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: $lim_{α \to 0^{+}} \frac{1}{α} cos (\frac{π}{α}) =$ $\infty,$ actual: DNE

A graph of the function (1/alpha) * cos (pi / alpha), which oscillates gently until the interval [-.2, .2], where it oscillates rapidly, going to infinity and negative infinity as it approaches the y axis.

2-47 §2.2 source ↗

In the following exercises, consider the graph of the function $y = f (x)$ shown here. Which of the statements about $y = f (x)$ are true and which are false? Explain why a statement is false.

A graph of a piecewise function with three segments and a point. The first segment is a curve opening upward with vertex at (-8, -6). This vertex is an open circle, and there is a closed circle instead at (-8, -3). The segment ends at (-2,3), where there is a closed circle. The second segment stretches up asymptotically to infinity along x=-2, changes direction to increasing at about (0,1.25), increases until about (2.25, 3), and decreases until (6,2), where there is an open circle. The last segment starts at (6,5), increases slightly, and then decreases into quadrant four, crossing the x axis at (10,0). All of the changes in direction are smooth curves.

$lim_{x \to {−2}^{+}} f (x) =$ $3$

Answer

False;

lim_{x \to {−2}^{+}} f (x) =

+ \infty

2-49 §2.2 source ↗

In the following exercises, consider the graph of the function $y = f (x)$ shown here. Which of the statements about $y = f (x)$ are true and which are false? Explain why a statement is false.

$lim_{x \to 6} f (x) =$ $5$

Answer

False;

lim_{x \to 6} f (x)

DNE since

lim_{x \to 6^{-}} f (x) =

2

and

lim_{x \to 6^{+}} f (x) =

5 .

2-51 §2.2 source ↗

In the following exercises, use the following graph of the function $y = f (x)$ to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first segment exists for x ≤1, and the second segment exists for x greater than 1. The first segment is linear with a slope of 1 and goes through the origin. Its endpoint is a closed circle at (1,1). The second segment is also linear with a slope of -1. It begins with the open circle at (1,2).

$lim_{x \to 1^{+}} f (x)$

Answer

2-53 §2.2 source ↗

In the following exercises, use the following graph of the function $y = f (x)$ to find the values, if possible. Estimate when necessary.

$lim_{x \to 2} f (x)$

Answer

2-55 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first is a linear function for x less than 0. There is an open circle at (0,1), and its slope is -1. The second segment is the right half of a parabola opening upward. Its vertex is a closed circle at (0, -4), and it goes through the point (2,0).

$lim_{x \to 0^{-}} f (x)$

Answer

2-57 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to 0} f (x)$

Answer

DNE

2-59 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with three segments, all linear. The first exists for x less than -2, has a slope of 1, and ends at the open circle at (-2, 0). The second exists over the interval [-2, 2], has a slope of -1, goes through the origin, and has closed circles at its endpoints (-2, 2) and (2,-2). The third exists for x greater than 2, has a slope of 1, and begins at the open circle (2,2).

$lim_{x \to {−2}^{-}} f (x)$

Answer

2-61 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to −2} f (x)$

Answer

DNE

2-63 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to 2^{+}} f (x)$

Answer

2-65 §2.2 source ↗

In the following exercises, use the graph of the function $y = g (x)$ shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first exists for x≥0 and is the left half of an upward opening parabola with vertex at the closed circle (0,3). The second exists for x greater than 0 and is the right half of a downward opening parabola with vertex at the open circle (0,0).

$lim_{x \to 0^{-}} g (x)$

Answer

2-67 §2.2 source ↗

In the following exercises, use the graph of the function $y = g (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to 0} g (x)$

Answer

DNE

2-69 §2.2 source ↗

In the following exercises, use the graph of the function $y = h (x)$ shown here to find the values, if possible. Estimate when necessary.

A graph of a function with two curves approaching 0 from quadrant 1 and quadrant 3. The curve in quadrant one appears to be the top half of a parabola opening to the right of the y axis along the x axis with vertex at the origin. The curve in quadrant three appears to be the left half of a parabola opening downward with vertex at the origin.

$lim_{x \to 0^{+}} h (x)$

Answer

2-71 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

A graph with a curve and a point. The point is a closed circle at (0,-2). The curve is part of an upward opening parabola with vertex at (1,-1). It exists for x greater than 0, and there is a closed circle at the origin.

$lim_{x \to 0^{-}} f (x)$

Answer

−2

2-73 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to 0} f (x)$

Answer

DNE

2-75 §2.2 source ↗

In the following exercises, use the graph of the function $y = f (x)$ shown here to find the values, if possible. Estimate when necessary.

$lim_{x \to 2} f (x)$

Answer

2-77 §2.2 source ↗

In the following exercises, sketch the graph of a function with the given properties.

$A s x \to$ $-$ $\infty,$ $f (x) \to$ $0,$ $lim_{x \to {−1}^{-}} f (x) =$ $− \infty,$ $lim_{x \to {−1}^{+}} f (x) =$ $\infty,$ $lim_{x \to 0} f (x) =$ $f (0),$ $f (0) =$ $1,$ $A s x \to$ $\infty,$ $f (x) \to$ $− \infty$

Answer

Answers may vary.

A graph of a piecewise function with two segments. The first segment is in quadrant three and asymptotically goes to negative infinity along the y axis and 0 along the x axis. The second segment consists of two curves. The first appears to be the left half of an upward opening parabola with vertex at (0,1). The second appears to be the right half of a downward opening parabola with vertex at (0,1) as well.

2-79 §2.2 source ↗

In the following exercises, sketch the graph of a function with the given properties.

$A s x \to$ $-$ $\infty,$ $f (x) \to$ $2,$ $lim_{x \to −2} f (x) =$ $− \infty,$ $A s x \to$ $\infty,$ $f (x) \to$ $2,$ $f (0) =$ $0$

Answer

Answers may vary.

A graph containing two curves. The first goes to 2 asymptotically along y=2 and to negative infinity along x = -2. The second goes to negative infinity along x=-2 and to 2 along y=2.

2-81 §2.2 source ↗

In the following exercises, sketch the graph of a function with the given properties.

Shock waves arise in many physical applications, ranging from supernovas to detonation waves. A graph of the density of a shock wave with respect to distance, x, is shown here. We are mainly interested in the location of the front of the shock, labeled $x_{SF}$ in the diagram.

A graph in quadrant one of the density of a shockwave with three labeled points: p1 and p2 on the y axis, with p1 greater than p2, and xsf on the x axis. It consists of y= p1 from 0 to xsf, x = xsf from y= p1 to y=p2, and y=p2 for values greater than or equal to xsf.

a. Evaluate $lim_{x \to x_{S F}^{+}} ρ (x) .$

b. Evaluate $lim_{x \to x_{S F}^{-}} ρ (x) .$

c. Evaluate $lim_{x \to x_{S F}} ρ (x) .$ Explain the physical meanings behind your answers.

Answer

ρ_{2}

ρ_{1}

c. DNE unless

ρ_{1} = ρ_{2} .

As you approach

x_{SF}

from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the “shock” yet and are at a lower density.

Checkpoint 2.11 §2.3 source ↗

Use the limit laws to evaluate

lim_{x \to 6} (2 x - 1) \sqrt{x + 4} .

In each step, indicate the limit law applied.

Answer

11 \sqrt{10}

Checkpoint 2.12 §2.3 source ↗

Evaluate

lim_{x \to −2} (3 x^{3} - 2 x + 7) .

Answer

−13;

Checkpoint 2.13 §2.3 source ↗

Evaluate

lim_{x \to −3} \frac{x^{2} + 4 x + 3}{x^{2} - 9} .

Answer

\frac{1}{3}

Checkpoint 2.14 §2.3 source ↗

Evaluate

lim_{x \to 5} \frac{\sqrt{x - 1} - 2}{x - 5} .

Answer

\frac{1}{4}

Checkpoint 2.15 §2.3 source ↗

Evaluate

lim_{x \to −3} \frac{\frac{1}{x + 2} + 1}{x + 3} .

Answer

−1;

Checkpoint 2.16 §2.3 source ↗

Evaluate

lim_{x \to 3} (\frac{1}{x - 3} - \frac{4}{x^{2} - 2 x - 3}) .

Answer

\frac{1}{4}

Checkpoint 2.17 §2.3 source ↗

Graph

f (x) = {\begin{matrix} - x - 2 if x < − 1 \\ 2 if x = −1 \\ x^{3} if x > − 1 \end{matrix}

and evaluate

lim_{x \to {−1}^{-}} f (x) .

Answer

The graph of a piecewise function with three segments. The first is a linear function, -x-2, for x less than -1. The x intercept is at (-2,0), and there is an open circle at (-1,-1). The next segment is simply the point (-1, 2). The third segment is the function x^3 for x greater than -1, which crossed the x axis and y axis at the origin.

$lim_{x \to {−1}^{-}} f (x) =$ $−1$

Checkpoint 2.18 §2.3 source ↗

Evaluate

lim_{x \to 1} \frac{x + 2}{{(x - 1)}^{2}} .

Answer

+∞

Checkpoint 2.19 §2.3 source ↗

Use the squeeze theorem to evaluate

lim_{x \to 0} x^{2} sin \frac{1}{x} .

Answer

Checkpoint 2.20 §2.3 source ↗

Evaluate

lim_{θ \to 0} \frac{1 - cos θ}{sin θ} .

Answer

2-83 §2.3 source ↗

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

$lim_{x \to 0} (4 x^{2} - 2 x + 3)$

Answer

Use constant multiple law and difference law:

lim_{x \to 0} (4 x^{2} - 2 x + 3) =

4 lim_{x \to 0} x^{2} -

2 lim_{x \to 0} x +

lim_{x \to 0} 3 =

3

2-85 §2.3 source ↗

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

$lim_{x \to −2} \sqrt{x^{2} - 6 x + 3}$

Answer

Use root law:

lim_{x \to −2} \sqrt{x^{2} - 6 x + 3} =

\sqrt{lim_{x \to −2} (x^{2} - 6 x + 3)} =

\sqrt{19}

2-87 §2.3 source ↗

In the following exercises, use direct substitution to evaluate each limit.

$lim_{x \to 7} x^{2}$

Answer

2-89 §2.3 source ↗

In the following exercises, use direct substitution to evaluate each limit.

$lim_{x \to 0} \frac{1}{1 + sin x}$

Answer

2-91 §2.3 source ↗

In the following exercises, use direct substitution to evaluate each limit.

$lim_{x \to 1} \frac{2 - 7 x}{x + 6}$

Answer

- \frac{5}{7}

2-93 §2.3 source ↗

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $0 / 0 .$ Then, evaluate the limit.

$lim_{x \to 4} \frac{x^{2} - 16}{x - 4}$

Answer

lim_{x \to 4} \frac{x^{2} - 16}{x - 4} =

\frac{16 - 16}{4 - 4} =

\frac{0}{0};

then,

lim_{x \to 4} \frac{x^{2} - 16}{x - 4} =

lim_{x \to 4} \frac{(x + 4) (x - 4)}{x - 4} =

8

2-95 §2.3 source ↗

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $0 / 0 .$ Then, evaluate the limit.

$lim_{x \to 6} \frac{3 x - 18}{2 x - 12}$

Answer

lim_{x \to 6} \frac{3 x - 18}{2 x - 12} =

\frac{18 - 18}{12 - 12} =

\frac{0}{0};

then,

lim_{x \to 6} \frac{3 x - 18}{2 x - 12} =

lim_{x \to 6} \frac{3 (x - 6)}{2 (x - 6)} =

\frac{3}{2}

2-97 §2.3 source ↗

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $0 / 0 .$ Then, evaluate the limit.

$lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3}$

Answer

lim_{x \to 9} \frac{t - 9}{\sqrt{t} - 3} =

\frac{9 - 9}{3 - 3} =

\frac{0}{0};

then,

lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3} =

lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3} \frac{\sqrt{t} + 3}{\sqrt{t} + 3} =

lim_{t \to 9} (\sqrt{t} + 3) =

6

2-99 §2.3 source ↗

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $0 / 0 .$ Then, evaluate the limit.

$lim_{θ \to π} \frac{sin θ}{tan θ}$

Answer

lim_{θ \to π} \frac{sin θ}{tan θ} =

\frac{sin π}{tan π} =

\frac{0}{0};

then,

lim_{θ \to π} \frac{sin θ}{tan θ} =

lim_{θ \to π} \frac{sin θ}{\frac{sin θ}{cos θ}} =

lim_{θ \to π} cos θ =

−1

2-101 §2.3 source ↗

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $0 / 0 .$ Then, evaluate the limit.

$lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1}$

Answer

lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1} =

\frac{\frac{1}{2} + \frac{3}{2} - 2}{1 - 1} =

\frac{0}{0};

then,

lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1} =

lim_{x \to 1 / 2} \frac{(2 x - 1) (x + 2)}{2 x - 1} =

\frac{5}{2}

2-103 §2.3 source ↗

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.

$lim_{x \to {−2}^{-}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}$

Answer

−∞

2-105 §2.3 source ↗

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.

$lim_{x \to 1^{-}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}$

Answer

−∞

2-107 §2.3 source ↗

In the following exercises, assume that $lim_{x \to 6} f (x) =$ $4,$ $lim_{x \to 6} g (x) =$ $9,$ and $lim_{x \to 6} h (x) =$ $6 .$ Use these three facts and the limit laws to evaluate each limit.

$lim_{x \to 6} 2 f (x) g (x)$

Answer

lim_{x \to 6} 2 f (x) g (x) =

2 lim_{x \to 6} f (x) lim_{x \to 6} g (x) =

72

2-109 §2.3 source ↗

In the following exercises, assume that $lim_{x \to 6} f (x) =$ $4,$ $lim_{x \to 6} g (x) =$ $9,$ and $lim_{x \to 6} h (x) =$ $6 .$ Use these three facts and the limit laws to evaluate each limit.

$lim_{x \to 6} (f (x) + \frac{1}{3} g (x))$

Answer

lim_{x \to 6} (f (x) + \frac{1}{3} g (x)) =

lim_{x \to 6} f (x) +

\frac{1}{3} lim_{x \to 6} g (x) =

7

2-111 §2.3 source ↗

In the following exercises, assume that $lim_{x \to 6} f (x) =$ $4,$ $lim_{x \to 6} g (x) =$ $9,$ and $lim_{x \to 6} h (x) =$ $6 .$ Use these three facts and the limit laws to evaluate each limit.

$lim_{x \to 6} \sqrt{g (x) - f (x)}$

Answer

lim_{x \to 6} \sqrt{g (x) - f (x)} =

\sqrt{lim_{x \to 6} g (x) - lim_{x \to 6} f (x)} =

\sqrt{5}

2-113 §2.3 source ↗

In the following exercises, assume that $lim_{x \to 6} f (x) =$ $4,$ $lim_{x \to 6} g (x) =$ $9,$ and $lim_{x \to 6} h (x) =$ $6 .$ Use these three facts and the limit laws to evaluate each limit.

$lim_{x \to 6} [(x + 1) \cdot f (x)]$

Answer

lim_{x \to 6} [(x + 1) f (x)] =

(lim_{x \to 6} (x + 1)) (lim_{x \to 6} f (x)) =

28

2-115 §2.3 source ↗

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

$f (x) = {\begin{cases} x^{2}, & x \leq 3 \\ x + 4, & x > 3 \end{cases}$

a. $lim_{x \to 3^{-}} f (x)$

b. $lim_{x \to 3^{+}} f (x)$

Answer

The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x≤3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x greater than 3. There is an open circle at (3, 7), and the slope is 1.

a. 9; b. 7

2-117 §2.3 source ↗

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

$h (x) = {\begin{cases} x^{2} - 2 x + 1, & x < 2 \\ 3 - x, & x \geq 2 \end{cases}$

a. $lim_{x \to 2^{-}} h (x)$

b. $lim_{x \to 2^{+}} h (x)$

Answer

The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x less than 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x≥ 2. It has a slope of -1 and an x intercept at (3,0).

a. 1; b. 1

2-119 §2.3 source ↗

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

Two graphs of piecewise functions. The upper is f(x), which has two linear segments. The first is a line with negative slope existing for x less than -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x greater than -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x less than -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4).

$lim_{x \to {−3}^{-}} (f (x) - 3 g (x))$

Answer

lim_{x \to {−3}^{-}} (f (x) - 3 g (x)) =

lim_{x \to {−3}^{-}} f (x) -

3 lim_{x \to {−3}^{-}} g (x) =

0 +

6 =

6

2-121 §2.3 source ↗

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

$lim_{x \to −5} \frac{2 + g (x)}{f (x)}$

Answer

lim_{x \to −5} \frac{2 + g (x)}{f (x)} =

\frac{2 + (lim_{x \to −5} g (x))}{lim_{x \to −5} f (x)} =

\frac{2 + 0}{2} =

1

2-123 §2.3 source ↗

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

$lim_{x \to 1} \sqrt[3]{f (x) - g (x)}$

Answer

lim_{x \to 1} \sqrt[3]{f (x) - g (x)} =

\sqrt[3]{lim_{x \to 1} f (x) - lim_{x \to 1} g (x)} =

\sqrt[3]{2 + 5} =

\sqrt[3]{7}

2-125 §2.3 source ↗

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

$lim_{x \to −9} [x \cdot f (x) + 2 \cdot g (x)]$

Answer

lim_{x \to −9} (x f (x) + 2 g (x)) =

(lim_{x \to −9} x) (lim_{x \to −9} f (x)) +

2 lim_{x \to −9} (g (x)) =

(−9) (6) +

2 (4) =

−46

2-127 §2.3 source ↗

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions $f (x), g (x),$ and $h (x)$ when possible.

[T] $lim_{θ \to 0} θ^{2} cos (\frac{1}{θ})$

Answer

The limit is zero.

The graph of three functions over the domain [-1,1], colored red, green, and blue as follows: red: theta^2, green: theta^2 * cos (1/theta), and blue: - (theta^2). The red and blue functions open upwards and downwards respectively as parabolas with vertices at the origin. The green function is trapped between the two.

2-129 §2.3 source ↗

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions $f (x), g (x),$ and $h (x)$ when possible.

[T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: $E (r) =$ $\frac{q}{4 π ε_{0} r^{2}},$ where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and $\frac{1}{4 π ε_{0}}$ is Coulomb’s constant: $8.988 \times 10^{9} N \cdot m^{2} / C^{2} .$

a. Use a graphing calculator to graph $E (r)$ given that the charge of the particle is $q = 10^{−10} .$

b. Evaluate $lim_{r \to 0^{+}} E (r) .$ What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

Answer

A graph of a function with two curves. The first is in quadrant two and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to negative infinity. The second is in quadrant one and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to infinity.

b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

Checkpoint 2.21 §2.4 source ↗

Using the definition, determine whether the function

f (x) = {\begin{matrix} 2 x + 1 & if x < 1 \\ 2 & if x = 1 \\ - x + 4 & if x > 1 \end{matrix}

is continuous at

x = 1 .

If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.

Answer

f is not continuous at 1 because

f (1) =

2 \neq 3 =

lim_{x \to 1} f (x) .

Checkpoint 2.22 §2.4 source ↗

For what values of x is

f (x) =

3 x^{4} -

4 x^{2}

continuous?

Answer

f (x)

is continuous at every real number.

Checkpoint 2.23 §2.4 source ↗

For

f (x) = {\begin{matrix} x^{2} & if x \neq 1 \\ 3 & if x = 1 \end{matrix},

decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

Answer

Discontinuous at 1; removable

Checkpoint 2.24 §2.4 source ↗

State the interval(s) over which the function

f (x) = \sqrt{x + 3}

is continuous.

Answer

[−3, + \infty)

Checkpoint 2.25 §2.4 source ↗

Evaluate

lim_{x \to π} sin (x - π) .

Answer

Checkpoint 2.26 §2.4 source ↗

Show that

f (x) =

x^{3} -

x^{2} -

3 x +

1

has a zero over the interval

[0, 1] .

Answer

f (0) =

1 >

0,

f (1) =

−2 <

0;

f (x)

is continuous over

[0, 1] .

It must have a zero on this interval.

2-131 §2.4 source ↗

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

$f (x) = \frac{1}{\sqrt{x}}$

Answer

The function is defined for all x in the interval

(0, \infty) .

2-133 §2.4 source ↗

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

$f (x) = \frac{x}{x^{2} - x}$

Answer

Removable discontinuity at

x = 0;

infinite discontinuity at

x = 1

2-135 §2.4 source ↗

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

$f (x) = \frac{5}{e^{x} - 2}$

Answer

Infinite discontinuity at

x = ln 2

2-137 §2.4 source ↗

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

$H (x) = tan 2 x$

Answer

Infinite discontinuities at

x = \frac{(2 k + 1) π}{4},

for

k =

0,

\pm 1,

\pm 2,

\pm 3,…

2-139 §2.4 source ↗

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

$f (x) =$ $\frac{2 x^{2} - 5 x + 3}{x - 1}$ at $x = 1$

Answer

No. It is a removable discontinuity.

2-141 §2.4 source ↗

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

$g (u) = {\begin{cases} \frac{6 u^{2} + u - 2}{2 u - 1} & if u \neq \frac{1}{2} \\ \frac{7}{2} & if u = \frac{1}{2} \end{cases},$ at $u = \frac{1}{2}$

Answer

Yes. It is continuous.

2-143 §2.4 source ↗

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

$f (x) = {\begin{cases} x^{2} - e^{x} & if x < 0 \\ x - 1 & if x \geq 0 \end{cases},$ at $x = 0$

Answer

Yes. It is continuous.

2-145 §2.4 source ↗

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

$f (x) = {\begin{matrix} 3 x + 2, & x < k \\ 2 x - 3, & k \leq x \leq 8 \end{matrix}$

Answer

k = −5

2-147 §2.4 source ↗

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

$f (x) = {\begin{matrix} \frac{x^{2} + 3 x + 2}{x + 2}, & x \neq - 2 \\ k, & x = −2 \end{matrix}$

Answer

k = −1

2-149 §2.4 source ↗

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

$f (x) = {\begin{matrix} \sqrt{k x}, & 0 \leq x \leq 3 \\ x + 1, & 3 < x \leq 10 \end{matrix}$

Answer

k = \frac{16}{3}

2-151 §2.4 source ↗

In the following exercises, use the Intermediate Value Theorem (IVT).

A particle moving along a line has at each time t a position function $s (t),$ which is continuous. Assume $s (2) = 5$ and $s (5) = 2 .$ Another particle moves such that its position is given by $h (t) =$ $s (t) -$ $t .$ Explain why there must be a value c for $2 < c < 5$ such that $h (c) = 0 .$

Answer

Since both s and

y = t

are continuous everywhere, then

h (t) = s (t) - t

is continuous everywhere and, in particular, it is continuous over the closed interval

[2, 5] .

Also,

h (2) = 3 > 0

and

h (5) =

−3 <

0 .

Therefore, by the IVT, there is a value

x = c

such that

h (c) = 0 .

2-153 §2.4 source ↗

In the following exercises, use the Intermediate Value Theorem (IVT).

Apply the IVT to determine whether $2^{x} = x^{3}$ has a solution in one of the intervals $[1.25, 1.375]$ or $[1.375, 1.5] .$ Briefly explain your response for each interval.

Answer

The function

f (x) = 2^{x} - x^{3}

is continuous over the interval

[1.25, 1.375]

and has opposite signs at the endpoints.

2-155 §2.4 source ↗

In the following exercises, use the Intermediate Value Theorem (IVT).

Let $f (x) = {\begin{matrix} 3 x, x > 1 \\ x^{3}, x < 1 \end{matrix} .$

a. Sketch the graph of f.

b. Is it possible to find a value k such that $f (1) = k,$ which makes $f (x)$ continuous for all real numbers? Briefly explain.

Answer

A graph of the given piecewise function containing two segments. The first, x^3, exists for x less than 1 and ends with an open circle at (1,1). The second, 3x, exists for x greater than 1. It beings with an open circle at (1,3).

b. It is not possible to redefine $f (1)$ since the discontinuity is a jump discontinuity.

2-157 §2.4 source ↗

In the following exercises, use the Intermediate Value Theorem (IVT).

Sketch the graph of a function $y = f (x)$ with properties i. through vi.

i. The domain of f is $(− \infty, + \infty) .$

ii. f has an infinite discontinuity at $x = −6 .$

iii. $f (−6) = 3$

iv. $lim_{x \to {−3}^{-}} f (x) =$ $lim_{x \to {−3}^{+}} f (x) =$ $2$

v. $f (−3) = 3$

vi. f is left continuous but not right continuous at $x = 3 .$

Answer

Answers may vary; see the following example:

A graph of a piecewise function with several segments. The first is an increasing line that exists for x less than -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 ≤ x less than -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 less than x ≤ 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x greater than 3.

2-159 §2.4 source ↗

In the following exercises, suppose $y = f (x)$ is defined for all x. For each description, sketch a graph with the indicated property.

Discontinuous at $x = 1$ with $lim_{x \to −1} f (x) =$ $−1$ and $lim_{x \to 2} f (x) =$ $4$

Answer

Answers may vary; see the following example:

The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x less than 1. It ends at (1,1). The second part is an increasing line that exists for x greater than 1. It begins at (1,3).

2-161 §2.4 source ↗

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

$f (t) =$ $\frac{2}{e^{t} - e^{- t}}$ is continuous everywhere.

Answer

False. It is continuous over

(− \infty, 0) \cup (0, \infty) .

2-163 §2.4 source ↗

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

If a function is not continuous at a point, then it is not defined at that point.

Answer

False. Consider

f (x) = {\begin{cases} x if x \neq 0 \\ 4 if x = 0 \end{cases} .

2-165 §2.4 source ↗

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

If $f (x)$ is continuous such that $f (a)$ and $f (b)$ have opposite signs, then $f (x) = 0$ has exactly one solution in $[a, b] .$

Answer

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider

f (x) = cos (x)

[- π, 2 π] .

2-167 §2.4 source ↗

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

If $f (x)$ is continuous everywhere and $f (a),$ $f (b) >$ $0,$ then there is no root of $f (x)$ in the interval $[a, b] .$

Answer

False. The IVT does not work in reverse! Consider

{(x - 1)}^{2}

over the interval

[−2, 2] .

2-169 §2.4 source ↗

[T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation $F (r) =$ $k_{e} \frac{| q_{1} q_{2} |}{r^{2}},$ where $k_{e}$ is Coulomb’s constant, $q_{i}$ are the magnitudes of the charges of the two particles, and r is the distance between the two particles.

Instead of making the force 0 at R, instead we let the force be 10⁻²⁰ for $r \geq R .$ Assume two protons, which have a magnitude of charge $1.6022 \times 10^{−19} C,$ and the Coulomb constant $k_{e} =$ $8.988 \times 10^{9} {Nm}^{2} / C^{2} .$ Is there a value R that can make this system continuous? If so, find it.

Answer

R =

0.0001519 m

2-171 §2.4 source ↗

Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by $F (d) =$ $-$ $m k / d^{2},$ where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant.

[T] After a certain distance D has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by $F (d) = {\begin{cases} - \frac{m k}{d^{2}} & if d < D \\ 10,000 & if d \geq D \end{cases} .$ Using the value of k found in the previous exercise, find the necessary condition D such that the force function remains continuous.

Answer

D = 345,826 km

2-173 §2.4 source ↗

Prove the following functions are continuous everywhere

$f (θ) = sin θ$

Answer

For all values of

a, f (a)

is defined,

lim_{θ \to a} f (θ)

exists, and

lim_{θ \to a} f (θ) =

f (a) .

Therefore,

f (θ)

is continuous everywhere.

2-175 §2.4 source ↗

Prove the following functions are continuous everywhere

Where is $f (x) = {\begin{cases} 0 if x is irrational \\ 1 if x is rational \end{cases}$ continuous?

Answer

Nowhere

Checkpoint 2.27 §2.5 source ↗

Complete the proof that $lim_{x \to 2} (3 x - 2) =$ $4$ by filling in the blanks.

Let _______.

Choose $δ =_______.$

Assume $0 <$ $| x -$ $____| <$ $____.$

Thus,

$|_______-$ $____| =$ $______________________________ε .$

Therefore, $lim_{x \to 2} (3 x - 2) =$ $4 .$

Answer

Let $ε > 0;$ choose $δ = \frac{ε}{3};$ assume $0 <$ $| x -$ $2 | <$ $δ .$

Thus, $| (3 x - 2) -$ $4 | =$ $| 3 x -$ $6 | =$ $| 3 | \cdot | x -$ $2 | <$ $3 \cdot δ =$ $3 \cdot (ε / 3) =$ $ε .$

Therefore, $lim_{x \to 2} 3 x -$ $2 =$ $4 .$

Checkpoint 2.28 §2.5 source ↗

Find δ corresponding to

ε > 0

for a proof that

lim_{x \to 9} \sqrt{x} = 3 .

Answer

Choose

δ =

min {9 - {(3 - ε)}^{2}, {(3 + ε)}^{2} - 9} .

Checkpoint 2.29 §2.5 source ↗

Complete the proof that $lim_{x \to 1} x^{2} = 1 .$

Let $ε > 0;$ choose $δ =$ $min {1, ε / 3};$ assume $0 <$ $| x -$ $1 | <$ $δ .$

Since $| x - 1 | < 1,$ we may conclude that $−1 <$ $x -$ $1 <$ $1 .$ Thus, $1 <$ $x +$ $1 <$ $3 .$ Hence, $| x + 1 | < 3 .$

Answer

| x^{2} -

1 | =

| x -

1 | \cdot | x +

1 | <

ε / 3 \cdot 3 =

ε

Checkpoint 2.30 §2.5 source ↗

Find

δ

corresponding to ε for a proof that

lim_{x \to 1^{-}} \sqrt{1 - x} =

0 .

Answer

δ = ε^{2}

2-177 §2.5 source ↗

In the following exercises, write the appropriate $ε$ - $δ$ definition for each of the given statements.

$lim_{t \to b} g (t) =$ $M$

Answer

For every

ε > 0,

there exists a

δ > 0,

so that if

0 <

| t -

b | <

δ,

then

| g (t) -

M | <

ε

2-179 §2.5 source ↗

In the following exercises, write the appropriate $ε$ - $δ$ definition for each of the given statements.

$lim_{x \to a} φ (x) =$ $A$

Answer

For every

ε > 0,

there exists a

δ > 0,

so that if

0 <

| x -

a | <

δ,

then

| φ (x) -

A | <

ε

2-181 §2.5 source ↗

The following graph of the function f satisfies $lim_{x \to 2} f (x) =$ $2 .$ In the following exercises, determine a value of $δ > 0$ that satisfies each statement.

A function drawn in quadrant one for x greater than 0. It is an increasing concave up function, with points approximately (0,0), (1, .5), (2,2), and (3,4).

If $0 <$ $| x -$ $2 | <$ $δ,$ then $| f (x) -$ $2 | <$ $0.5 .$

Answer

δ \leq 0.25

2-183 §2.5 source ↗

The following graph of the function f satisfies $lim_{x \to 3} f (x) =$ $−1 .$ In the following exercises, determine a value of $δ > 0$ that satisfies each statement.

A graph of a decreasing linear function, with points (0,2), (1,1), (2,0), (3,-1), (4,-2), and so on for x ≥ 0.

If $0 <$ $| x -$ $3 | <$ $δ,$ then $| f (x) +$ $1 | <$ $2 .$

Answer

δ \leq 2

2-185 §2.5 source ↗

The following graph of the function f satisfies $lim_{x \to 3} f (x) =$ $2 .$ In the following exercises, for each value of ε, find a value of $δ > 0$ such that the precise definition of limit holds true.

A graph of an increasing linear function intersecting the x axis at about (2.25, 0) and going through the points (3,2) and, approximately, (1,-5) and (4,5).

$ε = 3$

Answer

δ \leq 1

2-187 §2.5 source ↗

[T] In the following exercises, use a graphing calculator to find a number $δ$ such that the statements hold true.

$| \sqrt{x - 4} -$ $2 | <$ $0.1,$ $whenever | x -$ $8 | <$ $δ$

Answer

δ < 0.3900

2-189 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given limits.

$lim_{x \to 3} \frac{x^{2} - 9}{x - 3} =$ $6$

Answer

Let

δ = ε .

0 <

| x -

3 | <

ε,

then

| x +

3 -

6 | =

| x -

3 | <

ε .

2-191 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given limits.

$lim_{x \to 0} x^{4} = 0$

Answer

Let

δ = \sqrt[4]{ε} .

0 <

| x | <

\sqrt[4]{ε},

then

| x^{4} | =

x^{4} <

ε .

2-193 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given one-sided limits.

$lim_{x \to 5^{-}} \sqrt{5 - x} =$ $0$

Answer

Let

δ = ε^{2} .

5 -

ε^{2} <

x <

5,

then

| \sqrt{5 - x} | =

\sqrt{5 - x} <

ε .

2-195 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given one-sided limits.

$lim_{x \to 1^{-}} f (x) = 3, where f (x) = {\begin{matrix} 5 x - 2, if x < 1 \\ 7 x - 1, if x \geq 1 \end{matrix} .$

Answer

Let

δ = ε / 5 .

1 -

ε / 5 <

x <

1,

then

| f (x) -

3 | =

5 x -

5 <

ε .

2-197 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given infinite limits.

$lim_{x \to −1} \frac{3}{{(x + 1)}^{2}} =$ $\infty$

Answer

Let

δ = \sqrt{\frac{3}{M}} .

0 <

| x +

1 | <

\sqrt{\frac{3}{M}},

then

f (x) =

\frac{3}{{(x + 1)}^{2}} >

M .

2-199 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given infinite limits.

An engineer is using a machine to cut a flat square of Aerogel of area 144 cm². If there is a maximum error tolerance in the area of 8 cm², how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to $δ,$ ε, a, and L?

Answer

The engineer must cut within 0.328 cm of 12 cm on each side;

ε =

8,

δ =

0.328,

a =

12,

L =

144

2-201 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given infinite limits.

Using precise definitions of limits, prove that $lim_{x \to 0} f (x)$ does not exist, given that $f (x)$ is the ceiling function. (Hint: Try any $δ < 1 .)$

Answer

Answers may vary.

2-203 §2.5 source ↗

In the following exercises, use the precise definition of limit to prove the given infinite limits.

Using precise definitions of limits, determine $lim_{x \to 0} f (x)$ for $f (x) = {\begin{cases} x if x is rational \\ 0 if x is irrational \end{cases} .$ (Hint: Break into two cases, x rational and x irrational.)

Answer

2-205 §2.5 source ↗

For the following exercises, suppose that $lim_{x \to a} f (x) =$ $L$ and $lim_{x \to a} g (x) =$ $M$ both exist. Use the precise definition of limits to prove the following limit laws:

$lim_{x \to a} (f (x) + g (x)) =$ $L +$ $M$

Answer

\begin{array}{l} \lim_{x \to a} f x + \lim_{x \to a} g x \\ = L + M \end{array}

2-207 §2.5 source ↗

For the following exercises, suppose that $lim_{x \to a} f (x) =$ $L$ and $lim_{x \to a} g (x) =$ $M$ both exist. Use the precise definition of limits to prove the following limit laws:

$lim_{x \to a} [f (x) g (x)] =$ $L M .$ (Hint: $| f (x) g (x) -$ $L M | =$ $| f (x) g (x) -$ $f (x) M +$ $f (x) M -$ $L M | \leq | f (x) | | g (x) -$ $M | +$ $| M | | f (x) -$ $L | .)$

Answer

Answers may vary.

2-209 §2.R source ↗

True or False. In the following exercises, justify your answer with a proof or a counterexample.

You can use the quotient rule to evaluate $lim_{x \to 0} \frac{sin x}{x} .$

Answer

False

2-211 §2.R source ↗

True or False. In the following exercises, justify your answer with a proof or a counterexample.

If $lim_{x \to a} f (x)$ does not exist, then f is undefined at the point $x = a .$

Answer

False. A removable discontinuity is possible.

2-213 §2.R source ↗

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

$lim_{x \to 2} \frac{2 x^{2} - 3 x - 2}{x - 2}$

Answer

2-215 §2.R source ↗

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

$lim_{x \to 3} \frac{x^{3} - 2 x^{2} - 1}{3 x - 2}$

Answer

8 / 7

2-217 §2.R source ↗

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

$lim_{x \to −5} \frac{x^{2} + 25}{x + 5}$

Answer

DNE

2-219 §2.R source ↗

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

$lim_{x \to 1} \frac{x^{2} - 1}{x^{3} - 1}$

Answer

2 / 3

2-221 §2.R source ↗

In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

$lim_{x \to 4} \frac{4 - x}{\sqrt{x} - 2}$

Answer

−4;

2-223 §2.R source ↗

In the following exercises, use the squeeze theorem to prove the limit.

$lim_{x \to 0} x^{2} cos (2 π x) =$ $0$

Answer

Since

−1 \leq cos (2 π x) \leq 1,

then

-

x^{2} \leq x^{2} cos (2 π x) \leq x^{2} .

Since

lim_{x \to 0} x^{2} =

0 =

lim_{x \to 0} -

x^{2},

it follows that

lim_{x \to 0} x^{2} cos (2 π x) =

0 .

2-225 §2.R source ↗

In the following exercises, use the squeeze theorem to prove the limit.

Determine the domain such that the function $f (x) =$ $\sqrt{x - 2} +$ $x e^{x}$ is continuous over its domain.

Answer

[2, \infty]

2-227 §2.R source ↗

In the following exercises, determine the value of c such that the function remains continuous. Draw your resulting function to ensure it is continuous.

$f (x) = {\begin{cases} \sqrt{x + 1}, x > − 1 \\ x^{2} + c, x \leq - 1 \end{cases}$

Answer

c = −1

2-229 §2.R source ↗

In the following exercises, use the precise definition of limit to prove the limit.

$lim_{x \to 0} x^{3} = 0$

Answer

δ = \sqrt[3]{ε}

2-231 §2.R source ↗

In the following exercises, use the precise definition of limit to prove the limit.

A particle moving along a line has a displacement according to the function $x (t) =$ $t^{2} -$ $2 t +$ $4,$ where x is measured in meters and t is measured in seconds. Find the average velocity over the time period $t = [0, 2] .$

Answer

0 m / sec

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Problems and answers are adapted from Calculus Volume 1 by Gilbert Strang and Edwin “Jed” Herman, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.