NEW 2.3E: Limit Laws

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) \(lim_{x→0}(4x^2−2x+3)\)

Solution: Use constant multiple law and difference law:

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

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

3) \(lim_{x→−2}\sqrt{x^2−6x+3}\)

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

4) \(lim_{x→−1}(9x+1)^2\)

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

5) \(lim_{x→7}x^2)\)

Solution: 49

6) \(lim_{x→−2}(4x^2−1)\)

7) \(lim_{x→0}\frac{1}{1+sinx}\)

Solution: 1

8) \(lim_{x→2}e^{2x−x^2}\)

9) \(lim_{x→1}\frac{2−7x}{x+6}\)

Solution: \(−\frac{5}{7}\)

10) \(lim_{x→3}lne^{3x}\)

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

11) \(lim_{x→4}\frac{x^2−16}{x−4}\)

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

12) \(lim_{x→2}\frac{x−2}{x^2−2x}\)

13) \(lim_{x→6}\frac{3x−18}{2x−12}\)

Solution: \(lim_{x→6}\frac{3x−18}{2x−12}=\frac{18−18}{12−12}=\frac{0}{0}; then, lim_{x→6}\frac{3x−18}{2x− 12}=lim_{x→6}\frac{3(x−6)}{2(x−6)}=\frac{3}{2}\)

14) \(lim_{h→0}\frac{(1+h)^2−1}{h}\)

15) \(lim_{t→9}\frac{t−9}{\sqrt{t−3}}\)

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

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

17) \(lim_{θ→π}\frac{sinθ}{tanθ}\)

Solution: \(lim_{θ→π}\frac{sinθ}{tanθ}=\frac{sinπ}{tanπ}=\frac{0}{0}; then, lim_{θ→π}\frac{sinθ}{tanθ}=lim_{θ→ π}\frac{sinθ}{\frac{sinθ}{cosθ}}=lim_{θ→π}cosθ=−1\)

18) \(lim_{x→1}\frac{x^3−1}{x^2−1}\)

19) \(lim_{x→1/2}\frac{2x^2+3x−2}{2x−1}\)

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

20) \(lim_{x→−3}\frac{\sqrt{x+4}−1}{x+3}\)

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.

21) \(lim_{x→−2^−}\frac{2x^2+7x−4}{x^2+x−2}\)

Solution: −∞

22) \(lim_{x→−2^+}\frac{2x^2+7x−4}{x^2+x−2}\)

23) \(lim_{x→1^−}\frac{2x^2+7x−4}{x^2+x−2}\)

Solution: −∞

24) \(lim_{x→1^+}\frac{2x^2+7x−4}{x^2+x−2}\)

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

25) \(lim_{x→6}2f(x)g(x)\)

Solution: \(lim_{x→6}2f(x)g(x)=2lim_{x→6}f(x)lim_{x→6}g(x)=72\)

26) \(lim_{x→6}\frac{g(x)−1}{f(x)}\)

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

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

28) \(lim_{x→6}\frac{(h(x))^3}{2}\)

29) \(lim_{x→6}\sqrt{g(x)−f(x)}\)

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

30) \(lim_{x→6}x⋅h(x)\)

31) \(lim_{x→6}[(x+1)⋅f(x)]\)

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

32) \(lim_{x→6}(f(x)⋅g(x)−h(x))\)

[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) \(f(x)=\begin{cases}x2 & x≤3,\\ x+4 & x>3\end{cases}\)

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

Solution:

a. 9; b. 7

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

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

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

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

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

36) \(lim_{x→−3^+}(f(x)+g(x))\)

37) \(lim_{x→−3^−}(f(x)−3g(x))\)

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

38) \(lim_{x→0}\frac{f(x)g(x)}{3}\)

39) \(lim_{x→−5}\frac{2+g(x)}{f(x)}\)

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

40) \(lim_{x→1}(f(x))^2\)

41) \(lim_{x→1}\sqrt{f(x)−g(x)}\)

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

42) \(lim_{x→−7}(x⋅g(x))\)

43) \(lim_{x→−9}[x⋅f(x)+2⋅g(x)]\)

Solution: \(lim_{x→−9}(xf(x)+2g(x))=(lim_{x→−9}x)(lim_{x→−9}f(x))+2lim_{x→−9}(g(x))=(−9)(6)+2(4)=−46\)

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.

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

45) [T] \(lim_{θ→0}θ^2cos(\frac{1}{θ})

Solution: The limit is zero.

46) \(lim_{x→0}f(x)\), where \(f(x)=\begin{cases}0 & x rational\\ x^2 & x irrrational\end{cases}\)

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: \(E(r)=\frac{q}{4πε0_r^2}\), 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: \(8.988×109N⋅m^2/C^2\).

a. Use a graphing calculator to graph \(E(r)\) given that the charge of the particle is \(q=10^{−10}\).

b. Evaluate \(lim_{r→0^+}E(r)\). 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 \(lim_{x→0^+}V(\rho)\) and explain the physical meaning.

2.4: Continuity

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. \(lim_{x→−1}f(x)\)

b. \(lim_{x→1}f(x)\)

c. \(lim_{x→0^+}f(x)\)

d. \(lim_{x→2}f(x)\)

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

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

Solution: 5

214) \(lim_{x→0}3x^2−2x+4\)

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

Solution: 8/7

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

217) \(lim_{x→−5}\frac{x^2+25}{x+5}\)

Solution:DNE

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

219) \(lim_{x→1}\frac{x^2−1}{x^3−1}\)

Solution: 2/3

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

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

Solution: −4

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

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

223) \(lim_{x→0}x^2cos(2πx)=0\)

Solution: 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) \(lim_{x→0}x^3sin(\frac{π}{x})=0\)