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2.3E: Exercises

This page is a draft and is under active development. 

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Exercise 2.3E.1

In exercises 1 - 4, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

1) limx0(4x22x+3)

Answer

Use constant multiple law and difference law:

limx0(4x22x+3)=4limx0x22limx0x+limx03=0+0+3=3

2) limx1x3+3x2+547x

3) limx2x26x+3

Answer
Use root law: limx2x26x+3=limx2(x26x+3)=19

4) limx1(9x+1)2

Exercise 2.3E.2

In exercises 5 - 10, use direct substitution to evaluate the limit of each continuous function.

5) limx7x2

Answer
limx7x2=49

6) limx2(4x21)

7) limx011+sinx

Answer
limx011+sinx=1

8) limx2e2xx2

9) limx127xx+6

Answer
limx127xx+6=57

10) limx3lne3x

Exercise 2.3E.3

In exercises 11 - 20, use direct substitution to show that each limit leads to the indeterminate form 0/0. Then, evaluate the limit analytically.

11) limx4x216x4

Answer
When x=4,x216x4=161644=00;

then, limx4x216x4=limx4(x+4)(x4)x4=limx4(x+4)=4+4=8

12) limx2x2x22x

13) limx63x182x12

Answer
When x=6,3x182x12=18181212=00;

then, limx63x182x12=limx63(x6)2(x6)=limx632=32

14) limh0(1+h)21h

15) limt9t9t3

Answer
When t=9,t9t3=9933=00;

then, limt9t9t3=limt9t9t3t+3t+3=limt9(t9)(t+3)t9=limt9(t+3)=9+3=6

16) limh01a+h1ah, where a is a real-valued constant

17) \displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}

Answer
\displaystyle \text{When }θ = π, \quad\frac{\sin θ}{\tan θ}=\frac{\sin π}{\tan π}=\frac{0}{0};

then, \displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}=\lim_{θ→ π}\frac{\sin θ}{\frac{\sin θ}{\cos θ}}=\lim_{θ→π}\cos θ=\cos π=−1

18) \displaystyle \lim_{x→1}\frac{x^3−1}{x^2−1}

19) \displaystyle \lim_{x→1/2}\frac{2x^2+3x−2}{2x−1}

Answer
\displaystyle \text{When }x=1/2, \quad\frac{2x^2+3x−2}{2x−1}=\frac{\frac{1}{2}+\frac{3}{2}−2}{1−1}=\frac{0}{0};

then, \displaystyle \lim_{x→ 1/2}\frac{2x^2+3x−2}{2x−1}=\lim_{x→1/2}\frac{(2x−1)(x+2)}{2x−1}=\lim_{x→1/2}(x+2)=\frac{1}{2}+2=\frac{5}{2}

20) \displaystyle \lim_{x→−3}\frac{\sqrt{x+4}−1}{x+3}

Exercise \PageIndex{4}

In exercises 1 - 4, use direct substitution to obtain an undefined expression. Then, use the method used in Example 9 of this section to

In exercises 21 - 24, use direct substitution to obtain an undefined expression. Then, use the method used in Example 9 of this section to simplify the function and determine the limit.

21) \displaystyle \lim_{x→−2^−}\frac{2x^2+7x−4}{x^2+x−2}

Answer
−∞

22) \displaystyle \lim_{x→−2^+}\frac{2x^2+7x−4}{x^2+x−2}

23) \displaystyle \lim_{x→1^−}\frac{2x^2+7x−4}{x^2+x−2}

Answer
−∞

24) \displaystyle \lim_{x→1^+}\frac{2x^2+7x−4}{x^2+x−2}

Exercise \PageIndex{5}

In exercises 1 - 8, assume that \displaystyle \lim_{x→6}f(x)=4,\quad \lim_{x→6}g(x)=9, and \displaystyle \lim_{x→6}h(x)=6. Use these three

In exercises 25 - 32, assume that \displaystyle \lim_{x→6}f(x)=4,\quad \lim_{x→6}g(x)=9, and \displaystyle \lim_{x→6}h(x)=6. Use these three facts and the limit laws to evaluate each limit.

25) \displaystyle \lim_{x→6}2f(x)g(x)

Answer
\displaystyle \lim_{x→6}2f(x)g(x)=2\left(\lim_{x→6}f(x)\right)\left(\lim_{x→6}g(x)\right)=2 (4)(9)=72

26) \displaystyle \lim_{x→6}\frac{g(x)−1}{f(x)}

27) \displaystyle \lim_{x→6}\left(f(x)+\frac{1}{3}g(x)\right)

Answer
\displaystyle \lim_{x→6}\left(f(x)+\frac{1}{3}g(x)\right)=\lim_{x→6}f(x)+\frac{1}{3}\lim_{x→6}g(x)=4+\frac{1}{3}(9)=7

28) \displaystyle \lim_{x→6}\frac{\big(h(x)\big)^3}{2}

29) \displaystyle \lim_{x→6}\sqrt{g(x)−f(x)}

Answer
\displaystyle \lim_{x→6}\sqrt{g(x)−f(x)}=\sqrt{\lim_{x→6}g(x)−\lim_{x→6}f(x)}=\sqrt{9-4}=\sqrt{5}

30) \displaystyle \lim_{x→6}x⋅h(x)

31) \displaystyle \lim_{x→6}[(x+1)⋅f(x)]

Answer
\displaystyle \lim_{x→6}[(x+1)f(x)]=\left(\lim_{x→6}(x+1)\right)\left(\lim_{x→6}f(x)\right)=7(4)=28

32) \displaystyle \lim_{x→6}(f(x)⋅g(x)−h(x))

Exercise \PageIndex{6}

[T] In exercises 33 - 35, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

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

a. \displaystyle \lim_{x→3^−}f(x)

b. \displaystyle \lim_{x→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 > 3. There is an open circle at (3, 7), and the slope is 1.

a. 9; b. 7

34) g(x)=\begin{cases}x^3−1, & x≤0\\1, & x>0\end{cases}

a. \displaystyle \lim_{x→0^−}g(x)

b. \displaystyle \lim_{x→0^+}g(x)

35) h(x)=\begin{cases}x^2−2x+1, & x<2\\3−x, & x≥2\end{cases}

a. \displaystyle \lim_{x→2^−}h(x)

b. \displaystyle \lim_{x→2^+}h(x)

Exercise \PageIndex{7}

In exercises 36 - 43, 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 < -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 > -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 < -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).

36) \displaystyle \lim_{x→−3^+}(f(x)+g(x))

37) \displaystyle \lim_{x→−3^−}(f(x)−3g(x))

Answer
\displaystyle \lim_{x→−3^−}(f(x)−3g(x))=\lim_{x→−3^−}f(x)−3\lim_{x→−3^−}g(x)=0+6=6

38) \displaystyle \lim_{x→0}\frac{f(x)g(x)}{3}

39) \displaystyle \lim_{x→−5}\frac{2+g(x)}{f(x)}

Answer
\displaystyle \lim_{x→−5}\frac{2+g(x)}{f(x)}=\frac{2+\left(\displaystyle \lim_{x→−5}g(x)\right)}{\displaystyle \lim_{x→−5}f(x)}=\frac{2+0}{2}=1

40) \displaystyle \lim_{x→1}(f(x))^2

41) \displaystyle \lim_{x→1}\sqrt[3]{f(x)−g(x)}

Answer
\displaystyle \lim_{x→1}\sqrt[3]{f(x)−g(x)}=\sqrt[3]{\lim_{x→1}f(x)−\lim_{x→1}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}

42) \displaystyle \lim_{x→−7}(x⋅g(x))

43) \displaystyle \lim_{x→−9}[x⋅f(x)+2⋅g(x)]

Answer
\displaystyle \lim_{x→−9}(xf(x)+2g(x))=\left(\lim_{x→−9}x\right)\left(\lim_{x→−9}f(x)\right)+2\lim_{x→−9}g(x)=(−9)(6)+2(4)=−46
Exercise \PageIndex{8}

For exercises 44 - 46, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions f(x), g(x), and h(x) when possible.

44) [T] True or False? If 2x−1≤g(x)≤x^2−2x+3, then \displaystyle \lim_{x→2}g(x)=0.

45) [T] \displaystyle \lim_{θ→0}θ^2\cos\left(\frac{1}{θ}\right)

Answer

Since -1 \leq \cos(\dfrac{1}{\theta}) \leq 1 and \theta^2 \geq 0, -\theta^2 \leq \theta^2\cos(\dfrac{1}{\theta}) \leq \theta^2 . \\

Note that \lim_{\theta \to 0} \theta^2= - \lim_{\theta \to 0} \theta^2=0.

By squeeze theorem, \lim_{\theta \to 0} \theta^2\cos(\dfrac{1}{\theta})=0.

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.

46) \displaystyle \lim_{x→0}f(x), where f(x)=\begin{cases}0, & x\text{ rational}\\ x^2, & x\text{ irrrational}\end{cases}

Exercise \PageIndex{9}

47) [T] In physics, the magnitude of an electric field generated by a point charge at a distance r in a vacuum is governed by Coulomb’s law: E(r)=\dfrac{q}{4πε_0r^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 \dfrac{1}{4πε_0} is Coulomb’s constant: 8.988×109N⋅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 \displaystyle \lim_{r→0^+}E(r). What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

Answer

a.

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 the negative distance.

48) [T] The density of an object is given by its mass divided by its volume: ρ=m/V.

a. Use a calculator to plot the volume as a function of density (V=m/ρ), assuming you are examining something of mass 8 kg (m=8).

b. Evaluate \displaystyle \lim_{x→0^+}V(\rho) and explain the physical meaning.

Exercise \PageIndex{10}

Evaluate the following:

  1. \displaystyle \lim\limits_{x\to3}x^2-3x+7
  2. \displaystyle \lim\limits_{x\to\pi}\left ( \frac{x-3}{x+5}\right )^7
  3. \displaystyle \lim\limits_{x\to3}4^{{x^3}-8x}
  4. \displaystyle \lim\limits_{x\to0}\ln (1+x)
  5. \displaystyle \lim\limits_{x\to\pi}\frac{x^2+3x+5}{5x^2-2x-3}
  6. \displaystyle \lim\limits_{x\to\pi}\frac{3x+1}{1-x}
  7. \displaystyle \lim\limits_{x\to6}\frac{x^2-4x-12}{x^2-13x+42}
  8. \displaystyle \lim\limits_{x\to0}\frac{x^2+2x}{x^2-2x}
  9. \displaystyle \lim\limits_{x\to2}\frac{x^2+6x-16}{x^2-3x+2}
  10. \displaystyle \lim\limits_{x\to2}\frac{x^2-5x-14}{x^2+10x+16}
  11. \displaystyle \lim\limits_{x\to-2}\frac{x^2-5x-14}{x^2+10x+16}
  12. \displaystyle \lim\limits_{x\to-1}\frac{x^2+9x+8}{x^2-6x-7}
Answer

Under Construction

Exercise \PageIndex{11}

\displaystyle \lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}

Answer

\displaystyle \lim_{x \to 0} \frac{\sqrt{x+4}-2}{x} = \frac{\sqrt{0+4}-2}{0} =\left[\frac{0}{0}\right]

= \displaystyle \lim_{x \to 0} \frac{(\sqrt{x+4}-2) (\sqrt{x+4}+2)}{x (\sqrt{x+4}+2)}

= \displaystyle \lim_{x \to 0} \frac{((x+4)-4) }{x (\sqrt{x+4}+2)}

= \displaystyle \lim_{x \to 0} \frac{x }{x (\sqrt{x+4}+2)}

= \displaystyle \lim_{x \to 0} \frac{1 }{(\sqrt{x+4}+2)}= \frac{1 }{(\sqrt{0+4}+2)}= \frac{1 }{4}.

Exercise \PageIndex{12}

Evaluate the following limits:

1. \displaystyle \lim\limits_{x \to 4}\frac{1}{|4-x|}

Answer

\displaystyle \infty

2. \displaystyle \lim\limits_{x \to -1^-}\sqrt{1-x^2}

Answer

DNE

3. \displaystyle \lim\limits_{x \to -1^+}\sqrt{1-x^2}

Answer

0

4. \displaystyle \lim\limits_{x \to 2}\frac{|x-2|}{x^2+x-6}

Answer

\displaystyle \mbox{dne}

5. \displaystyle \lim\limits_{x \to 2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}

Answer

\displaystyle \frac{1}{4}

6. \displaystyle f(x)=\begin{cases} x^2 & \mbox{if } x \leq 1 \\ 2x & \mbox{if } x > 1\end{cases}

Answer

Under Construction

Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)


2.3E: Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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