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2.3E: Limit Laws and Techniques for Computing Limits EXERCISES

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Jan 17, 2020
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- 2.3E: Limit Laws and Techniques Exercises
- 2.4E: Infinite Limits EXERCISES

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2.3: The Limit Laws

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

83) $\displaystyle \lim_{x→0}(4x^2−2x+3)$

Answer:

Use constant multiple law and difference law:

$\displaystyle \lim_{x→0}(4x^2−2x+3)=4 \displaystyle \lim_{x→0}x^2−2 \displaystyle \lim{x→0}x+ \displaystyle \lim_{x→0}3=3$

84) $\displaystyle \lim_{x→1}\frac{x^3+3x^2+5}{4−7x}$

85) $\displaystyle \lim_{x→−2}\sqrt{x^2−6x+3}$

Answer:: Use root law: $\displaystyle \lim_{x→−2}\sqrt{x^2−6x+3}=\sqrt{ \displaystyle \lim_{x→−2}(x2−6x+3)}=\sqrt{19}$

86) $\displaystyle \lim_{x→−1}(9x+1)^2$

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

) $\displaystyle \lim_{x→7}x^2$

Answer:: 49

88) $\displaystyle \lim_{x→−2}(4x^2−1)$

) $\displaystyle \lim_{x→0}\frac{1}{1+\sin x}$

Answer:: 1

90) $\displaystyle \lim_{x→2}e^{2x−x^2}$

) $\displaystyle \lim_{x→1}\frac{2−7x}{x+6}$

Answer:: $−\frac{5}{7}$

92) $\displaystyle \lim_{x→3}\ln e^{3x}$

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

) $\displaystyle \lim_{x→4}\frac{x^2−16}{x−4}$

Answer:: $\displaystyle \lim_{x→4}\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0}; then, \displaystyle \lim_{x→4}\frac{x^2−16}{x−4}= \displaystyle \lim_{x→4}\frac{(x+4)(x−4)}{x−4}=8$

94) $\displaystyle \lim_{x→2}\frac{x−2}{x^2−2x}$

) $\displaystyle \lim_{x→6}\frac{3x−18}{2x−12}$

Answer:: $\displaystyle \lim_{x→6}\frac{3x−18}{2x−12}=\frac{18−18}{12−12}=\frac{0}{0}$

then, $\displaystyle \lim_{x→6}\frac{3x−18}{2x− 12}=\displaystyle \lim_{x→6}\frac{3(x−6)}{2(x−6)}=\frac{3}{2}$

96) $\displaystyle \lim_{h→0}\frac{(1+h)^2−1}{h}$

) $\displaystyle \lim _{t→9}\frac{t−9}{\sqrt{t}−3}$

Answer:: $\displaystyle \lim_{x→9}\frac{t−9}{\sqrt{t}−3}=\frac{9−9}{3−3}=\frac{0}{0}; then, \displaystyle \lim_{t→9}\frac{t−9}{\sqrt{t}−3} =\displaystyle \lim_{t→9}\frac{t−9}{\sqrt{t}−3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\displaystyle \lim_{t→9}(\sqrt{t}+3)=6$

) $\displaystyle \lim_{h→0}\frac{\frac{1}{a+h}−\frac{1}{a}}{h}$ , where a is a real-valued constant

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

Answer:: $\displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}=\frac{\sin π}{\tan π}=\frac{0}{0}; then, \displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}=\displaystyle \lim_{θ→ π}\frac{\sin θ}{\frac{\sin θ}{\cos θ}}=\displaystyle \lim_{θ→π}\cos θ=−1$

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

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

Answer:: $\displaystyle \lim_{x→1/2}\frac{2x^2+3x−2}{2x−1}=\frac{\frac{1}{2}+\frac{3}{2}−2}{1−1}=\frac{0}{0}; then, \displaystyle \lim_{x→ 1/2}\frac{2x^2+3x−2}{2x−1}=\displaystyle \lim_{x→1/2}frac{(2x−1)(x+2)}{2x−1}=\frac{5}{2}$

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

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

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

Answer:: −∞

104) $\displaystyle \lim _{x→−2^+}\frac{2x^2+7x−4}{x^2+x−2}$

105) $\displaystyle \lim _{x→1^−}\frac{2x^2+7x−4}{x^2+x−2}$

Answer:: −∞

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

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

) $\displaystyle \lim_{x→6}2f(x)g(x)$

Answer:: $\displaystyle \lim_{x→6}2f(x)g(x)=2\displaystyle \lim_{x→6}f(x)\displaystyle \lim_{x→6}g(x)=72$

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

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

Answer:: $\displaystyle \lim_{x→6}(f(x)+\frac{1}{3}g(x))=\displaystyle \lim_{x→6}f(x)+\frac{1}{3}\displaystyle \lim_{x→6}g(x)=7$ \

110) $\displaystyle \lim_{x→6}\frac{(h(x))^3}{2}$

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

Answer:: $\displaystyle \lim_{x→6}\sqrt{g(x)−f(x)}=\sqrt{\displaystyle \lim_{x→6}g(x)−\displaystyle \lim_{x→6}f(x)}=\sqrt{5}$

112) $\displaystyle \lim_{x→6}x⋅h(x)$

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

Answer:: $\displaystyle \lim_{x→6}[(x+1)f(x)]=(\displaystyle \lim_{x→6}(x+1))(\displaystyle \lim_{x→6}f(x))=28$

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

[T] In the following exercises, use the definition of the piecewise-defined function to evaluate the given limits (you may want to draw the graph).

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

a. $\displaystyle \lim_{x→3^−}f(x)$
b. $\displaystyle \lim_{x→3^+}f(x)$
c. $\displaystyle \lim_{x→3}f(x)$

Answer:

$CNX_Calc_Figure_02_03_202.jpeg$

a. 9; b. 7; c. DNE

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

a. $\displaystyle \lim_{x→0^−}g(x)$
b. $\displaystyle \lim_{x→0^+}g(x)$
c. $\displaystyle \lim_{x→0}g(x)$

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

a. $\displaystyle \lim_{x→2^−}h(x)$
b. $\displaystyle \lim_{x→2^+}h(x)$
c. $\displaystyle \lim_{x→2}h(x)$

Answer:: a. 1; b. 1; c. 1

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

$CNX_Calc_Figure_02_03_201.jpeg$

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

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

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

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

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

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

122) $\displaystyle \lim_{x→1}(f(x))^2$

123) $\displaystyle \lim_{x→1}\sqrt{f(x)−g(x)}$

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

124) $\displaystyle \lim_{x→−7}(x⋅g(x))$

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

Answer:: $\displaystyle \lim _{x→−9}(xf(x)+2g(x))=( \displaystyle \lim _{x→−9}x)( \displaystyle \lim _{x→−9}f(x))+2 \displaystyle \lim _{x→−9}(g(x))=(−9)(6)+2(4)=−46$

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] True or False? If $2x−1≤g(x)≤x^2−2x+3$ , then $\displaystyle \lim _{x→2}g(x)=0$ .

127) [T] $\displaystyle \lim _{θ→0}θ^2\cos(\frac{1}{θ})$

Answer:

The limit is zero.

$CNX_Calc_Figure_02_03_206.jpeg$

128) $\displaystyle \lim _{x→0}f(x)$ , where $f(x)=\begin{cases}0 & x rational\\ x^2 & x irrrational\end{cases}$

129) [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: $E(r)=\frac{q}{4πε0_r^2}$ , 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}{4πε_0} is Coulomb’s constant: $8.988×109\,\text{N⋅m}^2/\text{C}^2$ .

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:

$CNX_Calc_Figure_02_03_207.jpeg$

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.

130) [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.

Chapter Review Exercises

212) Using the graph, find each limit or explain why the limit does not exist.

a. $\displaystyle \lim_{x→−1}f(x)$

b. $\displaystyle \lim_{x→1}f(x)$

c. $\displaystyle \lim_{x→0^+}f(x)$

d. $\displaystyle \lim_{x→2}f(x)$

$CNX_Calc_Figure_02_05_207.jpeg$

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

213) $\displaystyle \lim_{x→2}\frac{2x^2−3x−2}{x−2}$

Answer:: 5

214) $\displaystyle \lim_{x→0}3x^2−2x+4$

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

Answer:: 8/7

216) $\displaystyle \lim_{x→π/2}\frac{cotx}{cosx}$ This is covered in section 2.4

217) $\displaystyle \lim_{x→−5}\frac{x^2+25}{x+5}$

Answer:: DNE

218) $\displaystyle \lim_{x→2}\frac{3x^2−2x−8}{x^2−4}$

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

Answer:: 2/3

220) $\displaystyle \lim_{x→1}\frac{x^2−1}{\sqrt{x}−1}$

221) \\displaystyle $lim_{x→4}\frac{4−x}{\sqrt{x}−2}$

Answer:: −4

222) $\displaystyle \lim_{x→4}\frac{1}{\sqrt{x}−2}$

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

223) $\displaystyle \lim_{x→0}x^2cos(2πx)=0$

Answer:: Since $−1≤cos(2πx)≤1$ , then $−x^2≤x^2cos(2πx)≤x^2$ . Since $lim_{x→0}x^2=0=lim_{x→0}−x^2$ , it follows that $lim_{x→0}x^2cos(2πx)=0$ .

224) $\displaystyle \lim_{x→0}x^3sin(\frac{π}{x})=0$

2.3: The Limit Laws

Chapter Review Exercises

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