2.3E: Limit Laws and Techniques for Computing 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) $$\displaystyle \lim_{x→0}(4x^2−2x+3)$$

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

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.

87) $$\displaystyle \lim_{x→7}x^2$$

49

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

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

1

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

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

$$−\frac{5}{7}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

−∞

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

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

−∞

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

107) $$\displaystyle \lim_{x→6}2f(x)g(x)$$

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

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

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

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

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

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

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

112) $$\displaystyle \lim_{x→6}x⋅h(x)$$

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

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

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).

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

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

a. 9; b. 7; c. DNE

.

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

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

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

1. a. $$\displaystyle \lim_{x→2^−}h(x)$$
2. b. $$\displaystyle \lim_{x→2^+}h(x)$$
3. c. $$\displaystyle \lim_{x→2}h(x)$$
a. 1; b. 1; c. 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

The limit is zero.

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?

a

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. $$\displaystyle \lim_{x→2}f(x)$$

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

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

5

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

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

8/7

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

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

DNE

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

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

2/3

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

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

−4

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

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

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

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