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Chapter 2: Limits and The Derivative

Chapter 2 Review Exercises: Limits

From Calculus, Volume 1, by Strang and Herman, OpenStax (Web), licensed under a CC BY-NC-SA 4.0 License.

True or false?

1.  A function has to be continuous at [latex]x=a[/latex] if the [latex]\underset{x\to a}{\lim}f(x)[/latex] exists.

2.  You can use the quotient rule to evaluate [latex]\underset{x\to 0}{\lim}\frac{\sin x}{x}[/latex].

Solution

False

3.  If there is a vertical asymptote at [latex]x=a[/latex] for the function [latex]f(x)[/latex], then [latex]f[/latex] is undefined at the point [latex]x=a[/latex].

4.  If [latex]\underset{x\to a}{\lim}f(x)[/latex] does not exist, then [latex]f[/latex] is undefined at the point [latex]x=a[/latex].

Solution

False. A removable discontinuity is possible.

5.  Using the graph of [latex]f(x)[/latex], find each of the following or explain why it does not exist.

  1. [latex]\underset{x\to -1^-}{\lim}f(x)[/latex]
  2. [latex]\underset{x\to -1^+}{\lim}f(x)[/latex]
  3. [latex]\underset{x\to -1}{\lim}f(x)[/latex]
  4. [latex]\underset{x\to 1}{\lim}f(x)[/latex]
  5. [latex]f(1)[/latex]
  6. [latex]\underset{x\to 0^+}{\lim}f(x)[/latex]
  7. [latex]\underset{x\to 0^-}{\lim}f(x)[/latex]
  8. [latex]\underset{x\to 0}{\lim}f(x)[/latex]
  9. [latex]\underset{x\to 2}{\lim}f(x)[/latex]

Graph with the x-axis ranging from -2 to 2 and the y-axis from -2 to 4. It features three distinct curves. The first curve is in the second quadrant, starting around (-1.5, 3) and ending at an open circle at (-1, 1). The second curve is in the first quadrant, beginning at an open circle at (0, 0) and ending at an open circle at (1, 1). The third curve is in the fourth quadrant, starting from a filled circle at (1, -1) and extending downward to about (0.5, -2).

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

6.  [latex]\underset{x\to 2}{\lim}\frac{2x^2-3x-2}{x-2}[/latex]

Solution

5

7.  [latex]\underset{x\to 0}{\lim}3x^2-2x+4[/latex]

8.  [latex]\underset{x\to 3}{\lim}\frac{x^3-2x^2-1}{3x-2}[/latex]

Solution

[latex]8/7[/latex]

9.  [latex]\underset{x\to -5}{\lim}\frac{x^2+25}{x+5}[/latex]

Solution

DNE

10.  [latex]\underset{x\to 2}{\lim}\frac{3x^2-2x-8}{x^2-4}[/latex]

11.  [latex]\underset{x\to 1}{\lim}\frac{x^2-1}{x^3-1}[/latex]

Solution

[latex]2/3[/latex]

12.  [latex]\underset{x\to 1}{\lim}\frac{x^2-1}{\sqrt{x}-1}[/latex]

13.  [latex]\underset{x\to 4}{\lim}\frac{4-x}{\sqrt{x}-2}[/latex]

Solution

−4

14.  [latex]\underset{x\to 4}{\lim}\frac{1}{\sqrt{x}-2}[/latex]

15. [latex]\underset{x\to 2^-}{\lim}\frac{\frac{1}{2}-\frac{1}{x}}{(x-2)^2}[/latex]

Solution

[latex]−\infty[/latex]

16. [latex]\underset{x\to 1^-}{\lim}\frac{|x-1|}{x^2-1}[/latex]

17. [latex]\underset{t\to 5}{\lim}\frac{t-5}{\sqrt{t-4}-1}[/latex]

Solution

2

18. [latex]\underset{x\to2}{\lim}\frac{x^2+x-6}{4-x^2}[/latex]

19. [latex]\underset{x\to 4}{\lim}\frac{\sqrt{8-x}-2}{4-x}[/latex]

Solution

[latex]\frac{1}{4}[/latex]

20. [latex]\underset{x\to 0}{\lim}\frac{\frac{1}{x+7}-\frac{1}{7}}{x}[/latex]

21. [latex]\underset{x\to -4^-}{\lim}\frac{\frac{1}{4} + \frac{1}{x}}{x+4}[/latex]

Solution

[latex]-\frac{1}{16}[/latex]

22. [latex]\underset{x\to 7^-}{\lim}\frac{-1}{(x-7)^{2021}}[/latex]

23. [latex]\underset{x\to -6^-}{\lim}\frac{2x+12}{|x+6|}[/latex]

Solution

[latex]-2[/latex]

24. [latex]\underset{x\to -6}{\lim}\frac{2x+12}{|x+6|}[/latex]

In the following exercises, evaluate the limits to infinity.

25. [latex]\underset{x\to -\infty}{\lim}\frac{\sqrt{11x^2+4x}}{5-4x}[/latex]

Solution

[latex]\frac{\sqrt{11}}{4}[/latex]

26. [latex]\underset{x\to -\infty}{\lim}\frac{\sqrt{7x^2-4x}}{2x-3}[/latex]

27. [latex]\underset{x\to \infty}{\lim}(x-\sqrt{x+1})[/latex]

Solution

[latex]\infty[/latex]

28. [latex]\underset{x\to \infty}{\lim}\sqrt{\frac{3x^2-1}{x+9x^2}}[/latex]

29. [latex]\underset{x\to \infty}{\lim}\sqrt{\frac{4x^2-1}{x+3x^2}}[/latex]

Solution

[latex]\frac{2}{\sqrt{3}}[/latex]

30. [latex]\underset{x\to -\infty}{\lim}\frac{3-x^2}{\sqrt[4]{x^8-4}}[/latex]

In the following exercises, determine the value of [latex]c[/latex] such that the function is continuous for the given value of [latex]x[/latex].

31. [latex]f(x)=\begin{cases} x^2+1 & \text{if} \, x < c \\ 2x & \text{if} \, x \le c \end{cases}[/latex]

32. [latex]f(x)=\begin{cases} \sqrt{x+1} & \text{if} \, x < -1 \\ x^2+c & \text{if} \, x \le -1 \end{cases}[/latex]

Solution

[latex]c=-1[/latex]

33. [latex]f(x)=\begin{cases} \frac{c^2}{3} (x+3) & \text{if} \, x < 0 \\ c+2 & \text{if} \, x = 0 \\ \frac{2}{3}(cx)^2 +1 & \text{if} \, x > 0 \end{cases}[/latex]

34. [latex]f(x)=\begin{cases} x+c & \text{if} \, x \le 3 \\ \frac{1}{x} & \text{if} \, x > 3\end{cases}[/latex]

Solution

[latex]c=-\frac{8}{3}[/latex]

In the following exercises, determine all horizontal and vertical asymptotes.

35. [latex]f(x) = \frac{3x^2-1}{2x^2+7x-4}[/latex]

36. [latex]f(x) = \frac{4x^3 -2x}{x^3-1}[/latex]

Solution

Horizontal: [latex]y=4[/latex]; Vertical: [latex]x = 1[/latex]

37. [latex]f(x) = \frac{\sqrt{2x^2+3}}{x^2-2x-3}[/latex]

37. [latex]f(x) = \frac{2x^2-1}{\sqrt{5x^4+2}}[/latex]

Solution

Horizontal: [latex]y=\frac{2}{\sqrt{5}}[/latex]; Vertical: none

In the following exercises, use the intermediate value theorem (IVT) to show that the given functions have an x-intercept in the given interval.

39. [latex]x^8 - 4x^3 = x + 1[/latex] on the interval [latex][-1,1][/latex]

40. [latex]f(x) =\frac{3}{x^4} -x^2+2[/latex] on the interval [latex][-2,-1][/latex]

Solution

Since [latex]f(x)[/latex] is continuous on [-2,-1] and [latex]f(-2) < 0[/latex] and [latex]f(-1) > 0[/latex], then by IVT, there exists a root on the given interval.

41. A ball is thrown into the air and the vertical position is given by [latex]x(t)=-4.9t^2+25t+5[/latex]. Use the intermediate value theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.

42. A particle moving along a line has a displacement according to the function [latex]x(t)=t^2-2t+4[/latex], where [latex]x[/latex] is measured in meters and [latex]t[/latex] is measured in seconds. Find the average velocity over the time period [latex]t=[0,2][/latex].

Solution

[latex]0[/latex] m/sec

In the following exercises, use the precise definition of limit to prove the limit.

43. [latex]\underset{x\to 1}{\lim}(8x+16)=24[/latex]

44. [latex]\underset{x\to 0}{\lim}x^3=0[/latex]

Solution

[latex]\delta =\sqrt[3]{\epsilon}[/latex]

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