# Continuity and One Sided Limits Worksheet

Continuity and One Sided Limits Worksheet
• Page 1
1.
If $f$($x$) = $\frac{1}{{x}^{2}+9}$, then which of the following statement(s) is/are correct?
I. $f$($x$) is continuous on the interval [- 1, 3]
II. $f$($x$) is not right continuous at - 1
III. $f$($x$) is continuous on the interval [- 4, 0]
IV. $f$($x$) is not left continuous at 0
 a. I, II only b. I only c. II only d. II, IV only e. I, III only

#### Solution:

limx-1+ f(x) = limx-1+ 1x2+9

= 1(-1)2+9 = 1 / 10 = f(- 1)

limx3- f(x) = limx3- 1x2+9

= 132+9 = 1 / 18 = f(3)

Draw the graph of f(x) in the interval [- 1, 3]

f(x) is continuous on the interval [- 1, 3].
[f(x) is right continuous at - 1 and left continuous at 3.]

limx- 4+ f(x) = limx- 4+ 1x2+9

= 1(-4)2+9 = 1 / 25 = f(- 4)

limx0- f(x) = limx0- 1x2+9

= 102+9 = 1 / 9 = f(0)

Draw the graph of f(x) in the interval [- 4, 0]

f(x) is continuous on the interval [- 4, 0].
[f(x) is right continuous at - 4 and left continuous at 0.]

The correct choice is E.

2.
Which of the following statements are correct for the function $f$($x$) = $x$2 - 3$x$ - 10 using the Intermediate Value Theorem?
I. $f$($x$) has a zero on [- 3, - 1]
II. $f$($x$) has no zero on [4, 6]
III. $f$($x$) has no zero on [-1, 0]
IV. $f$($x$) has no zero on [- 3, - 1]
 a. I, III only b. II, IV only c. IV only d. II, III only e. III, IV only

#### Solution:

f(x) = x2 - 3x - 10
[Write the function.]

f(x) is continuous on R.
[Polynomial functions are continuous on R.]

f(- 3) = 8, f(-1) = - 6, f(0) = - 10, f(4) = - 6 and f(6) = 8
[Find the values of f(-3), f(-1), f(0), f(4) and f(6).]

As f(- 3) > 0, f(-1) < 0, f(x) has a zero on [- 3, - 1]
[Use the Intermediate Value Theorem in [- 3, -1].]

As f(4) < 0, f(6) > 0, f(x) has a zero on [4, 6]
[Use the Intermediate Value Theorem in [4, 6].]

As f(-1) < 0, f(0) < 0, f(x) has no zero on [-1, 0]
[Use the Intermediate Value Theorem in [-1, 0].]

The correct choice is A.

3.
Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = | 2$x$ + 3 |.
I. $f$($x$) has a zero on [1, 2]
II. $f$($x$) has no zero on [- 3, - 2]
III. $f$($x$) has no zero on [- 2, - 1]
IV. $f$($x$) has a zero on [-1, 0]
 a. I, II only b. II, III only c. III, IV only d. II, IV only e. I, IV only

#### Solution:

f(x) = | 2x + 3 |
[Write the function.]

f(x) is continuous on R.

f(- 3) = 3, f(- 2) = 1, f(- 1) = 1, f(0) = 3, f(1) = 5, f(2) = 7
[Find the values of f(- 3), f(- 2), f(- 1), f(0), f(1) and f(2).]

As f(1) > 0, f(2) > 0, f(x) has no zero on [1, 2]
[Use Intermediate Value Theorem in [1, 2].]

As f(- 3) > 0, f(- 2) > 0, f(x) has no zero on [- 3, - 2]
[Use Intermediate Value Theorem in [- 3, - 2].]

As f(-2) > 0, f(-1) > 0, f(x) has no zero on [- 2, - 1]
[Use Intermediate Value Theorem in [- 2, - 1].]

As f(- 1) > 0, f(0) > 0, f(x) has no zero on [- 1, 0]
[Use Intermediate Value Theorem in [- 1, 0].]

The correct choice is B.

4.
Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = $\frac{2x}{{x}^{2}+4}$.
I. $f$($x$) has a zero on [- 1, 1]
II. $f$($x$) has no zero on [- 2, - 1]
III. $f$($x$) has a zero on [1, 2]
IV. $f$($x$) has a zero on [- 3, - 2]
 a. I, II only b. II, III only c. I, IV only d. II, IV only e. III, IV only

#### Solution:

f(x) = 2xx2+4
[Write the function.]

f(x) is continuous on R.

f(- 3) = - 6 / 13, f(- 2) = - 1 / 2, f(- 1) = - 2 / 5, f(1) = 2 / 5, f(2) = 1 / 2
[Find f(- 3), f(- 2), f(- 1), f(1) and f(2).]

As f(-1) < 0, f(1) > 0, f(x) has a zero on [-1, 1]
[Use Intermediate Value Theorem in [- 1, 1].]

As f(- 2) < 0, f(-1) < 0, f(x) has no zero on [- 2, -1]
[Use Intermediate Value Theorem in [- 2, -1].]

As f(1) > 0, f(2) > 0, f(x) has no zero on [1, 2]
[Use Intermediate Value Theorem in [1, 2].]

As f(- 3) < 0, f(- 2) < 0, f(x) has no zero on [- 3, - 2]
[Use Intermediate Value Theorem in [- 3, - 2].]

The correct choice is A.

5.
Use Intermediate Value Theorem to choose the correct statement from the following for the function $f$($x$) = $\frac{{4x}^{2}+3x}{x-1}$.
I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]
II. $f$($x$) has no zero on [- 2, -1]
III. $f$($x$) has a zero on [2, 3]
IV. $f$($x$) has a zero on [3, 4]
 a. I, II only b. II, III only c. III, IV only d. I, IV only e. I, III only

#### Solution:

f(x) = 4x2+3xx-1
[Write the function.]

f(x) is continuous on the interval (- ∞, 1) (1, ∞)

f(- 2) = -10 / 3, f(-1) = -1 / 2, f(-1 / 2) = 1 / 3, f(1 / 2) = - 5, f(2) = 22, f(3) = 45 / 2, f(4) = 76 / 3
[Find the values of f(- 2), f(-1), f(- 1 / 2), f(1 / 2), f(2), f(3) and f(4).]

As f(- 1 / 2) > 0, f(1 / 2) < 0, f(x) has a zero in [- 1 / 2, 1 / 2]
[Use Intermediate Value Theorem in [- 1 / 2, 1 / 2].]

As f(- 2) < 0, f(-1) < 0, f(x) has no zero in [- 2, -1]
[Use Intermediate Value Theorem in [- 2 , -1].]

As f(2) > 0, f(3) > 0, f(x) has no zero in [2, 3]
[Use Intermediate Value Theorem in [2, 3].]

As f(3) > 0, f(4) > 0, f(x) has no zero in [3, 4]
[Use Intermediate Value Theorem in [3, 4].]

The correct choice is A.

6.
Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = $e$2$x$ - 2$x$.
I. $f$($x$) has a zero on [- 2, - 1]
II. $f$($x$) has a zero on [0, $\frac{1}{2}$]
III. $f$($x$) has no zero on [$\frac{1}{2}$, 1]
IV. $f$($x$) has no zero on [- 1, 0]
 a. III, IV only b. I, III only c. I, II only d. II, IV only e. II, III only

#### Solution:

f(x) = e2x - 2x
[Write the function.]

f(x) is continuous for all real values of x.

f(- 2) = 4 + e- 4, f(- 1) = 2 + e- 2, f(0) = 1, f(1 / 2) = e - 1, f(1) = e2 - 2
[Find the values of f(- 2), f(-1), f(0), f(1 / 2) and f(1).]

As f(- 2) > 0, f(- 1) > 0, f(x) has no zero on [- 2, - 1]
[Use Intermediate Value Theorem in [- 2, - 1].]

As f(0) > 0, f(1 / 2) > 0, f(x) has no zero on [0, 1 / 2]
[Use Intermediate Value Theorem in [0, 1 / 2].]

As f(1 / 2) > 0, f(1) > 0, f(x) has no zero on [1 / 2, 1]
[Use Intermediate Value Theorem in [1 / 2, 1].]

As f(-1) > 0, f(0) > 0, f(x) has no zero on [- 1, 0]
[Use Intermediate Value Theorem in [- 1, 0].]

The correct choice is A.

7.
Use Intermediate Value Theorem to choose the correct statements for the function $f$($x$) = sin $x$.

 a. III, IV only b. II, IV only c. I, III only d. II, III only e. I, IV only

#### Solution:

f(x) = sin x
[Write the function.]

f(x) is continuous on R.

f(- π2) = - 1, f(- π4) = - 12, f(π6) = 1 / 2, f(π4) = 12, f(π3) = 32
[Find the values of f(- π2), f(- π4), f(π6), f(π4) and f(π3).]

As f(π4) > 0, f(π3) > 0, f(x) has no zero on [π4, π3]
[Use Intermediate Value Theorem in [π4, π3].]

As f(π3) > 0, f(π6) > 0, f(x) has no zero on [π3, π6]
[Use Intermediate Value Theorem in [π3, π6].]

As f(- π2) < 0, f(π4) > 0, f(x) has a zero on [- π2, π4]
[Use Intermediate Value Theorem in [- π2, π4].]

As f(- π2) < 0, f(- π4) < 0, f(x) has no zero on [- π2, - π4]
[Use Intermediate Value Theorem in [- π2, - π4].]

The correct choice is A.

8.
Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = - $x$ + ln($x$2 + 1).
I. $f$($x$) has a zero on [- 1, 1]
II. $f$($x$) has no zero on [1, 2]
III. $f$($x$) has a zero on [- 2, - 1]
Iv. $f$($x$) has a zero on [2, 3]
 a. I, II only b. II, III only c. II, IV only d. III, IV only e. I, IV only

#### Solution:

f(x) = - x + ln(x2 + 1)
[Write the function.]

f(x) is continuous on R.

f(- 2) = 2 + ln 5, f(- 1) = 1 + ln 2, f(1) = - 1 + ln 2, f(2) = - 2 + ln 5, f(3) = - 3 + ln 10
[Find the values of f(- 2), f(-1), f(1), f(2), f(3).]

As f(- 1) > 0, f(1) < 0, f(x) has a zero on [-1, 1]
[Use Intermediate Value Theorem in [- 1, 1].]

As f(1) < 0, f(2) < 0, f(x) has no zero on [1, 2]
[Use Intermediate Value Theorem in [1, 2].]

As f(- 2) > 0, f(- 1) > 0, f(x) has no zero on [- 2, -1]
[Use Intermediate Value Theorem in [- 2, - 1].]

As f(2) < 0, f(3) < 0, f(x) has no zero on [2, 3]
[Use Intermediate Value Theorem in [2, 3].]

The correct choice is A.

9.
Use Intermediate Value Theorem to choose the correct statements from the following for the function $f$($x$) = 2$x$ - $\sqrt{25-{x}^{2}}$.
I. $f$($x$) has a zero on [2, 3]
II. $f$($x$) has no zero on [1, 2]
III. $f$($x$) has a zero on [3, 4]
IV. $f$($x$) has a zero on [- 3, - 2]
 a. I, II only b. II, III only c. III, IV only d. II, IV only e. I, IV only

#### Solution:

f(x) = 2x - 25-x2
[Write the function.]

f(x) is continuous on [-5, 5].

f(- 3) = - 10, f(- 2) = - 4 - 21, f(1) = 2 - 26, f(2) = 4 - 21, f(3) = 2, f(4) = 5
[Find the values of f(- 3), f(- 2), f(1), f(2), f(3) and f(4).]

As f(2) < 0, f(3) > 0, f(x) has a zero on [2, 3]
[Use Intermediate Value Theorem in [2, 3].]

As f(1) < 0, f(2) < 0, f(x) has no zero on [1, 2]
[Use Intermediate Value Theorem in [1, 2].]

As f(3) > 0, f(4) > 0, f(x) has no zero on [3, 4]
[Use Intermediate Value Theorem in [3, 4].]

As f(- 3) < 0, f(- 2) < 0, f(x) has no zero on [- 3, - 2]
[Use Intermediate Value Theorem in [- 3, - 2].]

The correct choice is A.

10.
By using Intermediate Value Theorem, choose the correct statements from the following for the function $f$($x$) = 2$x$$\sqrt{{x}^{2}+5}$.
I. $f$($x$) has a zero on [- $\frac{1}{2}$, $\frac{1}{2}$]
II. $f$($x$) has a zero on [$\frac{1}{2}$, 1]
III. $f$($x$) has a zero on [1, 2]
IV. $f$($x$) has no zero on [- 2, - $\frac{1}{2}$]
 a. I, IV only b. II, IV only c. III, IV only d. I, III only e. II, III only

#### Solution:

f(x) = 2xx2+5
[Write the function.]

f(x) is continuous on R.

f(- 2) = - 12, f(- 1 / 2) = - 212, f(1 / 2) = 212, f(1) = 26, f(2) = 12
[Find the values of f(- 2), f(- 1 / 2), f(1 / 2), f(1) and f(2).]

As f(- 1 / 2) < 0, f(1 / 2) > 0, f(x) has a zero on [- 1 / 2, 1 / 2]
[Use Intermediate Value Theorem in [- 1 / 2, 1 / 2].]

As f(1 / 2) > 0, f(1) > 0, f(x) has no zero on [1 / 2 , 1]
[Use Intermediate Value Theorem in [1 / 2, 1].]

As f(1) > 0, f(2) > 0, f(x) has no zero on [1, 2]
[Use Intermediate Value Theorem in [1, 2].]

As f(- 2) < 0, f(- 1 / 2) < 0, f(x) has no zero on [- 2, - 1 / 2]
[Use Intermediate Value Theorem in [- 2, - 1 / 2].]

The correct choice is A.