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) limx→0(4x2−2x+3)
- Answer:
-
Use constant multiple law and difference law:
limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3
84) limx→1x3+3x2+54−7x
85) limx→−2√x2−6x+3
- Answer:
-
Use root law: limx→−2√x2−6x+3=√limx→−2(x2−6x+3)=√19
86) limx→−1(9x+1)2
In the following exercises, use direct substitution to evaluate each limit.
87) limx→7x2
- Answer:
- 49
88) limx→−2(4x2−1)
89) limx→011+sinx
- Answer:
- 1
90) limx→2e2x−x2
91) limx→12−7xx+6
- Answer:
- −57
92) limx→3lne3x
In the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0/0. Then, evaluate the limit.
93) limx→4x2−16x−4
- Answer:
- limx→4x2−16x−4=16−164−4=00;then,limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8
94) limx→2x−2x2−2x
95) limx→63x−182x−12
- Answer:
- limx→63x−182x−12=18−1812−12=00
then, limx→63x−182x−12=limx→63(x−6)2(x−6)=32
96) limh→0(1+h)2−1h
97) limt→9t−9√t−3
- Answer:
- limx→9t−9√t−3=9−93−3=00;then,limt→9t−9√t−3=limt→9t−9√t−3√t+3√t+3=limt→9(√t+3)=6
98) limh→01a+h−1ah, where a is a real-valued constant
99) limθ→πsinθtanθ
- Answer:
- limθ→πsinθtanθ=sinπtanπ=00;then,limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1
100) limx→1x3−1x2−1
101) limx→1/22x2+3x−22x−1
- Answer:
- limx→1/22x2+3x−22x−1=12+32−21−1=00;then,limx→1/22x2+3x−22x−1=limx→1/2frac(2x−1)(x+2)2x−1=52
102) limx→−3√x+4−1x+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) limx→−2−2x2+7x−4x2+x−2
- Answer:
- −∞
104) limx→−2+2x2+7x−4x2+x−2
105) limx→1−2x2+7x−4x2+x−2
- Answer:
- −∞
106) limx→1+2x2+7x−4x2+x−2
In the following exercises, assume that limx→6f(x)=4,limx→6g(x)=9, and limx→6h(x)=6. Use these three facts and the limit laws to evaluate each limit
107) limx→62f(x)g(x)
- Answer:
- limx→62f(x)g(x)=2limx→6f(x)limx→6g(x)=72
108) limx→6g(x)−1f(x)
109) limx→6(f(x)+13g(x))
- Answer:
- limx→6(f(x)+13g(x))=limx→6f(x)+13limx→6g(x)=7\
110) limx→6(h(x))32
111) limx→6√g(x)−f(x)
- Answer:
- limx→6√g(x)−f(x)=√limx→6g(x)−limx→6f(x)=√5
112) limx→6x⋅h(x)
113) limx→6[(x+1)⋅f(x)]
- Answer:
- limx→6[(x+1)f(x)]=(limx→6(x+1))(limx→6f(x))=28
114) limx→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)={x2x≤3,x+4x>3
- a. limx→3−f(x)
- b. limx→3+f(x)
- c. limx→3f(x)
- Answer:
-
a. 9; b. 7; c. DNE
.
116) g(x)={x3−1x≤01x>0
- a. limx→0−g(x)
- b. limx→0+g(x)
- c. limx→0g(x)
117) h(x)={x2−2x+1x<23−xx≥2
- a. limx→2−h(x)
- b. limx→2+h(x)
- c. limx→2h(x)
- Answer:
- a. 1; b. 1; c. 1
In the following exercises, use the following graphs and the limit laws to evaluate each limit.
118) limx→−3+(f(x)+g(x))
119) limx→−3−(f(x)−3g(x))
- Answer:
- limx→−3−(f(x)−3g(x))=limx→−3−f(x)−3limx→−3−g(x)=0+6=6
120) limx→0f(x)g(x)3
121) limx→−52+g(x)f(x)
- Answer:
- limx→−52+g(x)f(x)=2+(limx→−5g(x))limx→−5f(x)=2+02=1
122) limx→1(f(x))2
123) limx→1√f(x)−g(x)
- Answer:
- limx→13√f(x)−g(x)=3√limx→1f(x)−limx→1g(x)=3√2+5=3√7
124) limx→−7(x⋅g(x))
125) limx→−9[x⋅f(x)+2⋅g(x)]
- Answer:
- limx→−9(xf(x)+2g(x))=(limx→−9x)(limx→−9f(x))+2limx→−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)≤x2−2x+3, then limx→2g(x)=0.
127) [T] limθ→0θ2cos(1θ)
- Answer:
-
The limit is zero.
128) limx→0f(x), where f(x)={0xrationalx2xirrrational
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)=q4πε02r, 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: 8.988×109N⋅m2/C2.
a. Use a graphing calculator to graph E(r) given that the charge of the particle is q=10−10.
b. Evaluate limr→0+E(r). What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?
- Answer:
-
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 limx→0+V(ρ) and explain the physical meaning.
Chapter Review Exercises
212) Using the graph, find each limit or explain why the limit does not exist.
a. limx→−1f(x)
b. limx→1f(x)
c. limx→0+f(x)
d. limx→2f(x)
In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.
213) limx→22x2−3x−2x−2
- Answer:
- 5
214) limx→03x2−2x+4
215) limx→3x3−2x2−13x−2
- Answer:
- 8/7
216) limx→π/2cotxcosx This is covered in section 2.4
217) limx→−5x2+25x+5
- Answer:
- DNE
218) limx→23x2−2x−8x2−4
219) limx→1x2−1x3−1
- Answer:
- 2/3
220) limx→1x2−1√x−1
221) \\displaystyle limx→44−x√x−2
- Answer:
- −4
222) limx→41√x−2
In the following exercises, use the squeeze theorem to prove the limit.
223) limx→0x2cos(2πx)=0
- Answer:
- Since −1≤cos(2πx)≤1, then −x2≤x2cos(2πx)≤x2. Since limx→0x2=0=limx→0−x2, it follows that limx→0x2cos(2πx)=0.
224) limx→0x3sin(πx)=0