NEW 2.3E: Limit Laws
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- Jun 6, 2019
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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).
1) limx→0(4x2−2x+3)
Solution: Use constant multiple law and difference law:
limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3
2) limx→1x3+3x2+54−7x
3) limx→−2√x2−6x+3
Solution: Use root law: limx→−2√x2−6x+3=√limx→−2(x2−6x+3)=√19
4) limx→−1(9x+1)2
In the following exercises, use direct substitution to evaluate each limit.
5) limx→7x2)
Solution: 49
6) limx→−2(4x2−1)
7) limx→011+sinx
Solution: 1
8) limx→2e2x−x2
9) limx→12−7xx+6
Solution: −57
10) 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.
11) limx→4x2−16x−4
Solution:limx→4x2−16x−4=16−164−4=00;then,limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8
12) limx→2x−2x2−2x
13) limx→63x−182x−12
Solution: limx→63x−182x−12=18−1812−12=00;then,limx→63x−182x−12=limx→63(x−6)2(x−6)=32
14) limh→0(1+h)2−1h
15) limt→9t−9√t−3
Solution: 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
16) limh→01a+h−1ah, where a is a real-valued constant
17) limθ→πsinθtanθ
Solution: limθ→πsinθtanθ=sinπtanπ=00;then,limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1
18) limx→1x3−1x2−1
19) limx→1/22x2+3x−22x−1
Solution: 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
20) 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.
21) limx→−2−2x2+7x−4x2+x−2
Solution: −∞
22) limx→−2+2x2+7x−4x2+x−2
23) limx→1−2x2+7x−4x2+x−2
Solution: −∞
24) 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
25) limx→62f(x)g(x)
Solution: limx→62f(x)g(x)=2limx→6f(x)limx→6g(x)=72
26) limx→6g(x)−1f(x)
27) limx→6(f(x)+13g(x))
Solution: limx→6(f(x)+13g(x))=limx→6f(x)+13limx→6g(x)=7\
28) limx→6(h(x))32
29) limx→6√g(x)−f(x)
Solution: limx→6√g(x)−f(x)=√limx→6g(x)−limx→6f(x)=√5
30) limx→6x⋅h(x)
31) limx→6[(x+1)⋅f(x)]
Solution: limx→6[(x+1)f(x)]=(limx→6(x+1))(limx→6f(x))=28
32) limx→6(f(x)⋅g(x)−h(x))
[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.
33) f(x)={x2x≤3,x+4x>3
- a. limx→3−f(x)
- b. limx→3+f(x)
Solution:
a. 9; b. 7
34) g(x)={x3−1x≤01x>0
- a. limx→0−g(x)
- b. limx→0+g(x)
35) h(x)={x2−2x+1x<2x≥23−xx≥2
- a. limx→2−h(x)
- b. limx→2+h(x)
In the following exercises, use the following graphs and the limit laws to evaluate each limit.
36) limx→−3+(f(x)+g(x))
37) limx→−3−(f(x)−3g(x))
Solution: limx→−3−(f(x)−3g(x))=limx→−3−f(x)−3limx→−3−g(x)=0+6=6
38) limx→0f(x)g(x)3
39) limx→−52+g(x)f(x)
Solution: limx→−52+g(x)f(x)=2+(limx→−5g(x))limx→−5f(x)=2+02=1
40) limx→1(f(x))2
41) limx→1√f(x)−g(x)
Solution: limx→13√f(x)−g(x)=3√limx→1f(x)−limx→1g(x)=3√2+5=3√7
42) limx→−7(x⋅g(x))
43) limx→−9[x⋅f(x)+2⋅g(x)]
Solution: 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.
44) [T] True or False? If 2x−1≤g(x)≤x2−2x+3, then limx→2g(x)=0.
45) [T] \(lim_{θ→0}θ^2cos(\frac{1}{θ})
Solution: The limit is zero.
46) limx→0f(x), where f(x)={0xrationalx2xirrrational
47) [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?
Solution: 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.
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 limx→0+V(ρ) and explain the physical meaning.
Chapter Review Exercises
True or False. In the following exercises, justify your answer with a proof or a counterexample.
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
Solution: 5
214) limx→03x2−2x+4
215) limx→3x3−2x2−13x−2
Solution: 8/7
216) limx→π/2cotxcosx This is covered in section 2.4
217) limx→−5x2+25x+5
Solution:DNE
218) limx→23x2−2x−8x2−4
219) limx→1x2−1x3−1
Solution: 2/3
220) limx→1x2−1√x−1
221) limx→44−x√x−2
Solution: −4
222) limx→41√x−2
In the following exercises, use the squeeze theorem to prove the limit.
223) limx→0x2cos(2πx)=0
Solution: 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