2.3E: Exercises for Section 2.3

In exercises 1 - 4, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

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

Answer

Use constant multiple law, sum law, root law, and difference law:

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

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

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

Answer

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

4) \(\displaystyle \lim_{x→−1}(9x+1)^2\)

In exercises 5 - 10, use direct substitution to evaluate the limit of each continuous function.

5) \(\displaystyle \lim_{x→7}x^2\)

Answer

\(\displaystyle \lim_{x→7}x^2\;=\;49\)

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

7) \(\displaystyle \lim_{x→0}\frac{1}{1+\sin x}\)

Answer

\(\displaystyle \lim_{x→0}\frac{1}{1+\sin x}\;=\;1\)

8) \(\displaystyle \lim_{x→2}e^{2x−x^2}\)

9) \(\displaystyle \lim_{x→1}\frac{2−7x}{x+6}\)

Answer

\(\displaystyle \lim_{x→1}\frac{2−7x}{x+6}\;=\;−\frac{5}{7}\)

10) \(\displaystyle \lim_{x→3}\ln e^{3x}\)

In exercises 11 - 20, use direct substitution to show that each limit leads to the indeterminate form \(0/0\). Then, evaluate the limit analytically.

11) \(\displaystyle \lim_{x→4}\frac{x^2−16}{x−4}\)

Answer

\(\displaystyle \text{When }x = 4, \quad\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0};\)

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

12) \(\displaystyle \lim_{x→2}\frac{x−2}{x^2−2x}\)

13) \(\displaystyle \lim_{x→6}\frac{3x−18}{2x−12}\)

Answer

\(\displaystyle \text{When }x = 6, \quad\frac{3x−18}{2x−12}=\frac{18−18}{12−12}=\frac{0}{0};\)

then, \(\displaystyle \lim_{x→6}\frac{3x−18}{2x− 12}=\lim_{x→6}\frac{3(x−6)}{2(x−6)}=\lim_{x→6}\frac{3}{2}=\frac{3}{2}\)

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

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

Answer

\(\displaystyle \text{When }t = 9, \quad\frac{t−9}{\sqrt{t}−3}=\frac{9−9}{3−3}=\frac{0}{0};\)

then, \(\displaystyle \lim_{t→9}\frac{t−9}{\sqrt{t}−3} =\lim_{t→9}\frac{t−9}{\sqrt{t}−3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\lim_{t→9}\frac{(t−9)(\sqrt{t}+3)}{t - 9}=\lim_{t→9}(\sqrt{t}+3)=\sqrt{9}+3=6\)

16) \(\displaystyle \lim_{h→0}\frac{\dfrac{1}{a+h}−\dfrac{1}{a}}{h}\), where \(a\) is a real-valued constant and \(a\) is not zero. 

17) \(\displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}\)

Answer

\(\displaystyle \text{When }θ = π, \quad\frac{\sin θ}{\tan θ}=\frac{\sin π}{\tan π}=\frac{0}{0};\)

then, \(\displaystyle \lim_{θ→π}\frac{\sin θ}{\tan θ}=\lim_{θ→ π}\frac{\sin θ}{\frac{\sin θ}{\cos θ}}=\lim_{θ→π}\cos θ=\cos π=−1\)

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

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

Answer

\(\displaystyle \text{When }x=1/2, \quad\frac{2x^2+3x−2}{2x−1}=\frac{\frac{1}{2}+\frac{3}{2}−2}{1−1}=\frac{0}{0};\)

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

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

In exercises 21 - 24, use direct substitution to obtain an undefined expression. Then, use the method used in Example 9 of this section to simplify the function and determine the limit.

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

Answer

\(−∞\)

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

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

Answer

\(−∞\)

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

In exercises 25 - 32, assume that \(\displaystyle \lim_{x→6}f(x)=4,\quad \lim_{x→6}g(x)=9\), and \(\displaystyle \lim_{x→6}h(x)=6\). Use these three facts and the limit laws to evaluate each limit.

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

Answer

\(\displaystyle \lim_{x→6}2f(x)g(x)=2\left(\lim_{x→6}f(x)\right)\left(\lim_{x→6}g(x)\right)=2 (4)(9)=72\)

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

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

Answer

\(\displaystyle \lim_{x→6}\left(f(x)+\frac{1}{3}g(x)\right)=\lim_{x→6}f(x)+\frac{1}{3}\lim_{x→6}g(x)=4+\frac{1}{3}(9)=7\)

28) \(\displaystyle \lim_{x→6}\frac{\big(h(x)\big)^3}{2}\)

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

Answer

\(\displaystyle \lim_{x→6}\sqrt{g(x)−f(x)}=\sqrt{\lim_{x→6}g(x)−\lim_{x→6}f(x)}=\sqrt{9-4}=\sqrt{5}\)

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

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

Answer

\(\displaystyle \lim_{x→6}[(x+1)f(x)]=\left(\lim_{x→6}(x+1)\right)\left(\lim_{x→6}f(x)\right)=7(4)=28\)

32) \(\displaystyle \lim_{x→6}(f(x)⋅g(x)−h(x))\)

In exercises 33 - 35, Graph of each piecewise-defined function and study the graph to evaluate the given limits.

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

a. \(\displaystyle \lim_{x→3^−}f(x)\)

b. \(\displaystyle \lim_{x→3^+}f(x)\)

Answer

 

a. \(9\); b.\( 7\)

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

a. \(\displaystyle \lim_{x→0^−}g(x)\)

b. \(\displaystyle \lim_{x→0^+}g(x)\)

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

a. \(\displaystyle \lim_{x→2^−}h(x)\)

b. \(\displaystyle \lim_{x→2^+}h(x)\)

In exercises 36 - 43, use the following graphs and the limit laws to evaluate each limit.

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

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

Answer

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

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

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

Answer

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

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

41) \(\displaystyle \lim_{x→1}\sqrt[3]{f(x)−g(x)}\)

Answer

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

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

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

Answer

\(\displaystyle \lim_{x→−9}(xf(x)+2g(x))=\left(\lim_{x→−9}x\right)\left(\lim_{x→−9}f(x)\right)+2\lim_{x→−9}g(x)=(−9)(6)+2(4)=−46\)

For exercises 44 - 46, evaluate the limit using the squeeze theorem.

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

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

Answer

The limit is zero.

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

---