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Study Guides > Calculus Volume 1

Chapter 2 Review Exercises

True or False. In the following exercises, justify your answer with a proof or a counterexample.

1. A function has to be continuous at x=ax=a if the lim⁡x→af(x)\underset{x\to a}{\lim}f(x) exists.

2. You can use the quotient rule to evaluate lim⁡x→0sin⁡xx\underset{x\to 0}{\lim}\frac{\sin x}{x}.

Answer:

False

3. If there is a vertical asymptote at x=ax=a for the function f(x)f(x), then ff is undefined at the point x=ax=a.

4. If lim⁡x→af(x)\underset{x\to a}{\lim}f(x) does not exist, then ff is undefined at the point x=ax=a.

Answer:

False. A removable discontinuity is possible.

5. Using the graph, find each limit or explain why the limit does not exist.

  1. lim⁡x→−1f(x)\underset{x\to -1}{\lim}f(x)
  2. lim⁡x→1f(x)\underset{x\to 1}{\lim}f(x)
  3. lim⁡x→0+f(x)\underset{x\to 0^+}{\lim}f(x)
  4. lim⁡x→2f(x)\underset{x\to 2}{\lim}f(x)
A graph of a piecewise function with several segments. The first is a decreasing concave up curve existing for x < -1. It ends at an open circle at (-1, 1). The second is an increasing linear function starting at (-1, -2) and ending at (0,-1). The third is an increasing concave down curve existing from an open circle at (0,0) to an open circle at (1,1). The fourth is a closed circle at (1,-1). The fifth is a line with no slope existing for x > 1, starting at the open circle at (1,1).

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

6. lim⁡x→22x2−3x−2x−2\underset{x\to 2}{\lim}\frac{2x^2-3x-2}{x-2}

Answer:

5

7. lim⁡x→03x2−2x+4\underset{x\to 0}{\lim}3x^2-2x+4

8. lim⁡x→3x3−2x2−13x−2\underset{x\to 3}{\lim}\frac{x^3-2x^2-1}{3x-2}

Answer:

8/78/7

9. lim⁡x→π/2cot⁡xcos⁡x\underset{x\to \pi/2}{\lim}\frac{\cot x}{\cos x}

10. lim⁡x→−5x2+25x+5\underset{x\to -5}{\lim}\frac{x^2+25}{x+5}

Answer:

DNE

11. lim⁡x→23x2−2x−8x2−4\underset{x\to 2}{\lim}\frac{3x^2-2x-8}{x^2-4}

12. lim⁡x→1x2−1x3−1\underset{x\to 1}{\lim}\frac{x^2-1}{x^3-1}

Answer:

2/32/3

13. lim⁡x→1x2−1x−1\underset{x\to 1}{\lim}\frac{x^2-1}{\sqrt{x}-1}

14. lim⁡x→44−xx−2\underset{x\to 4}{\lim}\frac{4-x}{\sqrt{x}-2}

Answer:

−4

15. lim⁡x→41x−2\underset{x\to 4}{\lim}\frac{1}{\sqrt{x}-2}

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

16. lim⁡x→0x2cos⁡(2πx)=0\underset{x\to 0}{\lim}x^2\cos(2\pi x)=0

Answer:

Since −1≤cos⁡(2πx)≤1-1\le \cos (2\pi x)\le 1, then −x2≤x2cos⁡(2πx)≤x2-x^2\le x^2\cos(2\pi x)\le x^2. Since lim⁡x→0x2=0=lim⁡x→0−x2\underset{x\to 0}{\lim}x^2=0=\underset{x\to 0}{\lim}-x^2, it follows that lim⁡x→0x2cos⁡(2πx)=0\underset{x\to 0}{\lim}x^2\cos(2\pi x)=0.

17. lim⁡x→0x3sin⁡(πx)=0\underset{x\to 0}{\lim}x^3\sin(\frac{\pi}{x})=0

18. Determine the domain such that the function f(x)=x−2+xexf(x)=\sqrt{x-2}+xe^x is continuous over its domain.

Answer:

[2,∞)[2,\infty)

In the following exercises, determine the value of cc such that the function remains continuous. Draw your resulting function to ensure it is continuous.

19. f(x)={x2+1if x>c2xif x≤cf(x)=\begin{cases} x^2+1 & \text{if} \, x>c \\ 2x & \text{if} \, x \le c \end{cases}

20. f(x)={x+1if x>−1x2+cif x≤−1f(x)=\begin{cases} \sqrt{x+1} & \text{if} \, x > -1 \\ x^2+c & \text{if} \, x \le -1 \end{cases}

Answer:

c=−1c=-1

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

21. lim⁡x→1(8x+16)=24\underset{x\to 1}{\lim}(8x+16)=24

22. lim⁡x→0x3=0\underset{x\to 0}{\lim}x^3=0

Answer:

δ=ϵ3\delta =\sqrt[3]{\epsilon}

23. A ball is thrown into the air and the vertical position is given by x(t)=−4.9t2+25t+5x(t)=-4.9t^2+25t+5. 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.

24. A particle moving along a line has a displacement according to the function x(t)=t2−2t+4x(t)=t^2-2t+4, where xx is measured in meters and tt is measured in seconds. Find the average velocity over the time period t=[0,2]t=[0,2].

Answer:

00 m/sec

25. From the previous exercises, estimate the instantaneous velocity at t=2t=2 by checking the average velocity within t=0.01t=0.01 sec.

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