Problem Set: The Limit Laws

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

1. $\underset{x\to 0}{\lim}(4x^2-2x+3)$

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Use constant multiple law and difference law: $\underset{x\to 0}{\lim}(4x^2-2x+3)=4\underset{x\to 0}{\lim}x^2-2\underset{x\to 0}{\lim}x+\underset{x\to 0}{\lim}3=3$

2. $\underset{x\to 1}{\lim}\frac{x^3+3x^2+5}{4-7x}$

3. $\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}$

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Use root law: $\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}=\sqrt{\underset{x\to -2}{\lim}(x^2-6x+3)}=\sqrt{19}$

4. $\underset{x\to -1}{\lim}(9x+1)^2$

In the following exercises (5-10), use direct substitution to evaluate each limit.

5. $\underset{x\to 7}{\lim}x^2$

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6. $\underset{x\to -2}{\lim}(4x^2-1)$

7. $\underset{x\to 0}{\lim}\dfrac{1}{1+ \sin x}$

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8. $\underset{x\to 2}{\lim}e^{2x-x^2}$

9. $\underset{x\to 1}{\lim}\dfrac{2-7x}{x+6}$

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$-\frac{5}{7}$

10. $\underset{x\to 3}{\lim}\ln e^{3x}$

In the following exercises (11-20), use direct substitution to show that each limit leads to the indeterminate form $\frac{0}{0}$ . Then, evaluate the limit.

11. $\underset{x\to 4}{\lim}\dfrac{x^2-16}{x-4}$

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$\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\frac{16-16}{4-4}=\frac{0}{0}$ ; then, $\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\underset{x\to 4}{\lim}\frac{(x+4)(x-4)}{x-4}=8$

12. $\underset{x\to 2}{\lim}\dfrac{x-2}{x^2-2x}$

13. $\underset{x\to 6}{\lim}\dfrac{3x-18}{2x-12}$

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$\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0}$ ; then,

$\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\underset{x\to 6}{\lim}\frac{3(x-6)}{2(x-6)}=\frac{3}{2}$

14. $\underset{h\to 0}{\lim}\dfrac{(1+h)^2-1}{h}$

15. $\underset{t\to 9}{\lim}\dfrac{t-9}{\sqrt{t}-3}$

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$\underset{t \to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\frac{9-9}{3-3}=\frac{0}{0}$ ; then, $\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to 9}{\lim}(\sqrt{t}+3)=6$

16. $\underset{h\to 0}{\lim}\dfrac{\frac{1}{a+h}-\frac{1}{a}}{h}$ , where $a$ is a real-valued constant

17. $\underset{\theta \to \pi}{\lim}\dfrac{\sin \theta}{\tan \theta}$

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$\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\frac{\sin \pi}{\tan \pi}=\frac{0}{0}$ ; then,

$\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}}=\underset{\theta \to \pi}{\lim}\cos \theta =-1$ .

18. $\underset{x\to 1}{\lim}\dfrac{x^3-1}{x^2-1}$

19. $\underset{x\to 1/2}{\lim}\dfrac{2x^2+3x-2}{2x-1}$

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$\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\frac{\frac{1}{2}+\frac{3}{2}-2}{1-1}=\frac{0}{0}$ ; then, $\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\underset{x\to 1/2}{\lim}\frac{(2x-1)(x+2)}{2x-1}=\frac{5}{2}$

20. $\underset{x\to -3}{\lim}\dfrac{\sqrt{x+4}-1}{x+3}$

In the following exercises (21-24), use direct substitution to obtain an undefined expression. Then, use the method of (Figure) to simplify the function to help determine the limit.

21. $\underset{x\to -2^-}{\lim}\dfrac{2x^2+7x-4}{x^2+x-2}$

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$-\infty$

22. $\underset{x\to -2^+}{\lim}\dfrac{2x^2+7x-4}{x^2+x-2}$

23. $\underset{x\to 1^-}{\lim}\dfrac{2x^2+7x-4}{x^2+x-2}$

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$-\infty$

24.

$\underset{x\to 1^+}{\lim}\dfrac{2x^2+7x-4}{x^2+x-2}$

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

25. $\underset{x\to 6}{\lim}2f(x)g(x)$

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$\underset{x\to 6}{\lim}2f(x)g(x)=2\underset{x\to 6}{\lim}f(x)\underset{x\to 6}{\lim}g(x)=72$

26. $\underset{x\to 6}{\lim}\dfrac{g(x)-1}{f(x)}$

27. $\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))$

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$\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))=\underset{x\to 6}{\lim}f(x)+\frac{1}{3}\underset{x\to 6}{\lim}g(x)=7$

28. $\underset{x\to 6}{\lim}\dfrac{(h(x))^3}{2}$

29. $\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}$

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$\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}=\sqrt{\underset{x\to 6}{\lim}g(x)-\underset{x\to 6}{\lim}f(x)}=\sqrt{5}$

30. $\underset{x\to 6}{\lim}x \cdot h(x)$

31. $\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]$

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$\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]=(\underset{x\to 6}{\lim}(x+1))(\underset{x\to 6}{\lim}f(x))=28$ .

32.

$\underset{x\to 6}{\lim}(f(x)\cdot g(x)-h(x))$

In the following 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. [T] $f(x)=\begin{cases} x^2, & x \le 3 \\ x+4, & x > 3 \end{cases}$

$\underset{x\to 3^-}{\lim}f(x)$
$\underset{x\to 3^+}{\lim}f(x)$

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34. [T] $g(x)=\begin{cases} x^3 - 1, & x \le 0 \\ 1, & x > 0 \end{cases}$

$\underset{x\to 0^-}{\lim}g(x)$
$\underset{x\to 0^+}{\lim}g(x)$

35. [T] $h(x)=\begin{cases} x^2-2x+1, & x < 2 \\ 3 - x, & x \ge 2 \end{cases}$

1. $\underset{x\to 2^-}{\lim}h(x)$
2. $\underset{x\to 2^+}{\lim}h(x)$

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

36. $\underset{x\to -3^+}{\lim}(f(x)+g(x))$

37. $\underset{x\to -3^-}{\lim}(f(x)-3g(x))$

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$\underset{x\to -3^-}{\lim}(f(x)-3g(x))=\underset{x\to -3^-}{\lim}f(x)-3\underset{x\to -3^-}{\lim}g(x)=0+6=6$

38. $\underset{x\to 0}{\lim}\dfrac{f(x)g(x)}{3}$

39. $\underset{x\to -5}{\lim}\dfrac{2+g(x)}{f(x)}$

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$\underset{x\to -5}{\lim}\frac{2+g(x)}{f(x)}=\frac{2+(\underset{x\to -5}{\lim}g(x))}{\underset{x\to -5}{\lim}f(x)}=\frac{2+0}{2}=1$

40. $\underset{x\to 1}{\lim}(f(x))^2$

41. $\underset{x\to 1}{\lim}\sqrt[3]{f(x)-g(x)}$

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$\underset{x\to 1}{\lim}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\underset{x\to 1}{\lim}f(x)-\underset{x\to 1}{\lim}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}$

42. $\underset{x\to -7}{\lim}(x \cdot g(x))$

43. $\underset{x\to -9}{\lim}[xf(x)+2g(x)]$

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$\underset{x\to -9}{\lim}(x\cdot f(x)+2g(x))=(\underset{x\to -9}{\lim}x)(\underset{x\to -9}{\lim}f(x))+2\underset{x\to -9}{\lim}(g(x))=(-9)(6)+2(4)=-46$

For the following problems (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 $2x-1\le g(x)\le x^2-2x+3$ , then $\underset{x\to 2}{\lim}g(x)=0$ .

45. [T] $\underset{\theta \to 0}{\lim}\theta^2 \cos\left(\frac{1}{\theta}\right)$

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46. $\underset{x\to 0}{\lim}f(x)$ , where $f(x)=\begin{cases} 0, & x \, \text{rational} \\ x^2, & x \, \text{irrational} \end{cases}$

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: $E(r)=\large \frac{q}{4\pi \varepsilon_0 r^2}$ , 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 $\large \frac{1}{4\pi \varepsilon_0}$ is Coulomb’s constant: $8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$ .

Use a graphing calculator to graph $E(r)$ given that the charge of the particle is $q=10^{-10}$ .
Evaluate $\underset{r\to 0^+}{\lim}E(r)$ . What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

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b. $\underset{r\to 0^+}{\lim}E(r)=\infty$ . 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: $\rho =\dfrac{m}{V}$ .

Use a calculator to plot the volume as a function of density $(V=\frac{m}{\rho})$ , assuming you are examining something of mass 8 kg ( $m=8$ ).
Evaluate $\underset{\rho \to 0^+}{\lim}V(\rho)$ and explain the physical meaning.

Limits: Supplemental Content

Problem Set: The Limit Laws

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