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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) limx0(4x22x+3)

Answer:

Use constant multiple law and difference law:

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

84) limx1x3+3x2+547x

85) limx2x26x+3

Answer:

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

86) limx1(9x+1)2

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

87) limx7x2

Answer:
49

88) limx2(4x21)

89) limx011+sinx

Answer:
1

90) limx2e2xx2

91) limx127xx+6

Answer:
57

92) limx3lne3x


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

93) limx4x216x4

Answer:
limx4x216x4=161644=00;then,limx4x216x4=limx4(x+4)(x4)x4=8

94) limx2x2x22x

95) limx63x182x12

Answer:
limx63x182x12=18181212=00

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

96) limh0(1+h)21h

97) limt9t9t3

Answer:
limx9t9t3=9933=00;then,limt9t9t3=limt9t9t3t+3t+3=limt9(t+3)=6

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

99) \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

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}

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}

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}

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

107) \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)}

109) \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}

111) \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).

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)
Answer:

CNX_Calc_Figure_02_03_202.jpeg

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)
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.

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})

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:

a

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


2.3E: Limit Laws and Techniques for Computing Limits EXERCISES is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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