# 2.4E: Exercises for Section 2.4

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

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

Use constant multiple 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) $$\displaystyle \lim_{x→−2}\sqrt{x^2−6x+3}$$

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) $$\displaystyle \lim_{x→7}x^2$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$−∞$$

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

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

$$−∞$$

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

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

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

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

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

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

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

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

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

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

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

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

[T] In exercises 33 - 35, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

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

a. $$\displaystyle \lim_{x→3^−}f(x)$$

b. $$\displaystyle \lim_{x→3^+}f(x)$$

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

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

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

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

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

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

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

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

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

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

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

For exercises 44 - 46, evaluate the limit using the squeeze theorem. 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 $$\displaystyle \lim_{x→2}g(x)=0$$.

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

The limit is zero.

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

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: $$E(r)=\dfrac{q}{4πε_0r^2}$$, 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 $$\dfrac{1}{4πε_0}$$ 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 $$\displaystyle \lim_{r→0^+}E(r)$$. What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

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 $$\displaystyle \lim_{x→0^+}V(\rho)$$ and explain the physical meaning.