15

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=a if the \underset{x\to a}{\lim}f(x) exists.

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

Solution

False

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

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

Solution

False. A removable discontinuity is possible.

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

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

Solution

5

7. \underset{x\to 0}{\lim}3x^2-2x+4

8. \underset{x\to 3}{\lim}\frac{x^3-2x^2-1}{3x-2}

Solution

8/7

9. \underset{x\to \pi/2}{\lim}\frac{\cot x}{\cos x}

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

Solution

DNE

11. \underset{x\to 2}{\lim}\frac{3x^2-2x-8}{x^2-4}

12. \underset{x\to 1}{\lim}\frac{x^2-1}{x^3-1}

Solution

2/3

13. \underset{x\to 1}{\lim}\frac{x^2-1}{\sqrt{x}-1}

14. \underset{x\to 4}{\lim}\frac{4-x}{\sqrt{x}-2}

Solution

−4

15. \underset{x\to 4}{\lim}\frac{1}{\sqrt{x}-2}

16. \underset{x\to 2^-}{\lim}\frac{\frac{1}{2}-\frac{1}{x}}{(x-2)^2}

Solution

−\infty

17. \underset{x\to 1^-}{\lim}\frac{|x-1|}{x^2-1}

18. \underset{t\to 5}{\lim}\frac{t-5}{\sqrt{t-4}-1}

Solution

2

19. \underset{x\to2}{\lim}\frac{x^2+x-6}{4-x^2}

20. \underset{x\to 4}{\lim}\frac{\sqrt{8-x}-2}{4-x}

Solution

 \frac{1}{4}

21. \underset{x\to 0}{\lim}\frac{\frac{1}{x+7}-\frac{1}{7}}{x}

22. \underset{x\to -4^-}{\lim}\frac{\frac{1}{4} + \frac{1}{x}}{x+4}

Solution

 -\frac{1}{16}

23. \underset{x\to 7^-}{\lim}\frac{-1}{(x-7)^{2021}}

24. \underset{x\to -6^-}{\lim}\frac{2x+12}{|x+6|}

Solution

 -2

25. \underset{x\to -6}{\lim}\frac{2x+12}{|x+6|}

In the following exercises, evaluate the limits to infinity.

26. \underset{x\to -\infty}{\lim}\frac{\sqrt{11x^2+4x}}{5-4x}

Solution

 \frac{\sqrt{11}}{4}

27. \underset{x\to -\infty}{\lim}\frac{\sqrt{7x^2-4x}}{2x-3}

28. \underset{x\to \infty}{\lim}(x-\sqrt{x+1})

Solution

 \infty

29. \underset{x\to \infty}{\lim}\sqrt{\frac{3x^2-1}{x+9x^2}}

30. \underset{x\to \infty}{\lim}\sqrt{\frac{4x^2-1}{x+3x^2}}

Solution

 \frac{2}{\sqrt{3}}

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

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

32. \underset{x\to 0}{\lim}x^2\cos(2\pi x)=0

Solution

Since -1\le \cos (2\pi x)\le 1, then -x^2\le x^2\cos(2\pi x)\le x^2. Since \underset{x\to 0}{\lim}x^2=0=\underset{x\to 0}{\lim}-x^2, it follows that \underset{x\to 0}{\lim}x^2\cos(2\pi x)=0.

33. \underset{x\to 0}{\lim}x^3\sin(\frac{\pi}{x})=0

34. \underset{x\to \infty }{\text{lim}} \frac{1}{x} \cos (2x)

Solution

Since -1\le \cos (2x)\le 1, then -\frac{1}{x}\le \frac{1}{x} \cos(2x)\le \frac{1}{x}. Since \underset{x\to \infty}{\lim}\frac{1}{x}=0=\underset{x\to \infty}{\lim}-\frac{1}{x}, it follows that \underset{x\to \infty}{\lim}\frac{1}{x} \cos(2x)=0.

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

35. f(x)=\begin{cases} x^2+1 & \text{if} \, x \symbol{"3E} c \\ 2x & \text{if} \, x \le c \end{cases}

36. f(x)=\begin{cases} \sqrt{x+1} & \text{if} \, x \symbol{"3E} -1 \\ x^2+c & \text{if} \, x \le -1 \end{cases}

Solution

c=-1

37. 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}

38. f(x)=\begin{cases} x+c & \text{if} \, x \le 3 \\ \frac{1}{x} & \text{if} \, x > 3\end{cases}

Solution

c=-\frac{8}{3}

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

39. f(x) = \frac{3x^2-1}{2x^2+7x-4}

40. f(x) = \frac{4x^3 -2x}{x^3-1}

Solution

Horizontal:  y=4, Vertical:  x = 1

41. f(x) = \frac{\sqrt{2x^2+3}}{x^2-2x-3}

42. f(x) = \frac{2x^2-1}{\sqrt{5x^4+2}}

Solution

Horizontal:  y=\frac{2}{\sqrt{5}}, Vertical: none

In the following exercises, use the Intermediate Value Theorem to show that the given functions have an x-intercept in the given interval.

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

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

Solution

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

45. A ball is thrown into the air and the vertical position is given by x(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.

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

Solution

0 m/sec

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

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

48. \underset{x\to 0}{\lim}x^3=0

Solution

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

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