Calculus 1 Quiz: Removing Discontinuities

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Removing DiscontinuitiesQuestion 1 of 20

The function $f(x) = \frac{e^{2x} - 1}{e^x - 1}$ has a removable discontinuity at $x=0$ . What value should be assigned to $f(0)$ to make the function continuous?

0

2

1

e

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Calculus 1 Quiz

Calculus 1 Quiz: Removing Discontinuities

Practice Removing Discontinuities in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Removing Discontinuities, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The function $f(x) = \frac{e^{2x} - 1}{e^x - 1}$ has a removable discontinuity at $x=0$ . What value should be assigned to $f(0)$ to make the function continuous?

$0$
$2$ (correct answer)
$1$
$e$

Explanation: When you encounter a function with a removable discontinuity, you're dealing with a limit problem where the function appears undefined at a point but actually approaches a specific value. The key is finding what value the function approaches as x gets arbitrarily close to the problematic point. At

x = 0

, this function gives us

\frac{e^0 - 1}{e^0 - 1} = \frac{0}{0}

, an indeterminate form. To find the limiting value, we can factor the numerator using the difference of squares:

e^{2x} - 1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)

. This gives us

f(x) = \frac{(e^x - 1)(e^x + 1)}{e^x - 1}

. For

x \neq 0

, we can cancel the

(e^x - 1)

terms, leaving

f(x) = e^x + 1

. As

x

approaches 0, this simplified expression approaches

e^0 + 1 = 1 + 1 = 2

. Therefore, setting

f(0) = 2

makes the function continuous. Choice A (0) likely comes from incorrectly thinking the

\frac{0}{0}

form equals zero. Choice C (1) might result from forgetting to add 1 after finding

e^0

. Choice D (e) could stem from misapplying exponential properties or incorrect algebraic manipulation. Study tip: When you see indeterminate forms like

\frac{0}{0}

, always look for algebraic simplification opportunities—factoring, rationalizing, or L'Hôpital's rule. The removable discontinuity's "repair value" is always the limit as x approaches the problematic point.

Question 2

The function $f(x) = \frac{x^3 + 27}{x^2 + 4x + 3}$ has a removable discontinuity. If a continuous function $g(x)$ is created by removing this discontinuity, what is the value of $g(-3)$ ?

$-\frac{27}{2}$ (correct answer)
$\frac{9}{2}$
$\frac{27}{4}$
$\frac{27}{2}$

Explanation: The function has discontinuities where the denominator is zero:

x^2 + 4x + 3 = (x+3)(x+1) = 0

, so at

x=-3

and

x=-1

. To check if the discontinuity at

x=-3

is removable, we factor the numerator using the sum of cubes formula,

a^3+b^3 = (a+b)(a^2-ab+b^2)

x^3+27 = (x+3)(x^2 - 3x + 9)

. So,

f(x) = \frac{(x+3)(x^2 - 3x + 9)}{(x+3)(x+1)}

. The

(x+3)

factor cancels, indicating a removable discontinuity at

x=-3

. The value of

g(-3)

is the limit:

\lim_{x \to -3} \frac{x^2 - 3x + 9}{x+1} = \frac{(-3)^2 - 3(-3) + 9}{-3+1} = \frac{9+9+9}{-2} = \frac{27}{-2} = -\frac{27}{2}

Question 3

For what value of the constant $k$ does the function $f(x) = \frac{x^2 + kx - 10}{x - 2}$ have a removable discontinuity at $x=2$ ?

$k = -5$
$k = -3$
$k = 3$ (correct answer)
$k = 7$

Explanation: For a rational function to have a removable discontinuity at

x=c

, the numerator must be zero at

x=c

, allowing the factor

(x-c)

to be cancelled. Thus, the numerator

x^2 + kx - 10

must be zero when

x=2

. Substituting

x=2

gives:

(2)^2 + k(2) - 10 = 0 \Rightarrow 4 + 2k - 10 = 0 \Rightarrow 2k = 6 \Rightarrow k = 3

. When

k=3

f(x) = \frac{x^2+3x-10}{x-2} = \frac{(x+5)(x-2)}{x-2}

, which simplifies to

x+5

for

x \neq 2

Question 4

Consider the function $f(x) = \frac{\ln(x)}{x-1}$ . A new function $g(x)$ is defined to be continuous for all $x>0$ by removing the discontinuity of $f(x)$ . What is the value of $g(1)$ ?

$0$
$\frac{1}{e}$
$1$ (correct answer)
$e$

Explanation: The function

f(x)

is undefined at

x=1

. To find the value that would make it continuous, we must calculate the limit as

x \to 1

. Direct substitution of

x=1

into

f(x)

results in the indeterminate form

\frac{\ln(1)}{1-1} = \frac{0}{0}

. We can apply L'Hôpital's Rule:

\lim_{x \to 1} \frac{\ln(x)}{x-1} = \lim_{x \to 1} \frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = \frac{1/1}{1} = 1

. Therefore, defining

g(1)=1

removes the discontinuity.

Question 5

Let $g(x) = \frac{x^2 - c^2}{|x|-c}$ for a constant $c > 0$ . The function has removable discontinuities at $x=c$ and $x=-c$ . Which of the following definitions for $h(x)$ creates a function that is continuous on $(-\infty, \infty)$ ?

(correct answer)

Explanation: To make the function continuous, we must define

h(c)

and

h(-c)

to be equal to their respective limits. For

x=c

: As

x \to c

, we can assume

x>0

, so

|x|=x

\lim_{x \to c} \frac{x^2-c^2}{x-c} = \lim_{x \to c} \frac{(x-c)(x+c)}{x-c} = \lim_{x \to c} (x+c) = 2c

. For

x=-c

: As

x \to -c

, we can assume

x<0

, so

|x|=-x

\lim_{x \to -c} \frac{x^2-c^2}{-x-c} = \lim_{x \to -c} \frac{(x-c)(x+c)}{-(x+c)} = \lim_{x \to -c} -(x-c) = -(-c-c) = 2c

. Since both limits are

2c

, we must define the function to be

2c

at both

x=c

and

x=-c

to remove the discontinuities.

Question 6

Let $f(x)$ be a function with a removable discontinuity at $x=c$ , and let $g(x)$ be a function that is continuous at $x=c$ . Let $h(x) = f(x)g(x)$ . Which condition on $g(x)$ is sufficient to guarantee that $h(x)$ is continuous at $x=c$ ?

$g(c) = \lim_{x \to c} f(x)$
$g(c) = 1$
$g(c) = f(c)$
$g(c) = 0$ (correct answer)

Explanation: When analyzing products of functions with discontinuities, you need to consider how the behavior of each function affects the overall continuity. Since

f(x)

has a removable discontinuity at

x=c

, we know that

\lim_{x \to c} f(x)

exists but either

f(c)

is undefined or

f(c) \neq \lim_{x \to c} f(x)

. For

h(x) = f(x)g(x)

to be continuous at

x=c

, we need

\lim_{x \to c} h(x) = h(c)

. Using the product rule for limits:

\lim_{x \to c} h(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) = \lim_{x \to c} f(x) \cdot g(c)

g(c) = 0

, then this limit equals

\lim_{x \to c} f(x) \cdot 0 = 0

. Also,

h(c) = f(c) \cdot g(c) = f(c) \cdot 0 = 0

(assuming

f(c)

is defined). Therefore,

\lim_{x \to c} h(x) = h(c) = 0

, making

h(x)

continuous at

x=c

. This works regardless of the specific values of

f(c)

\lim_{x \to c} f(x)

. Choice A sets an arbitrary relationship that doesn't ensure continuity of the product. Choice B (

g(c) = 1

) would make

\lim_{x \to c} h(x) = \lim_{x \to c} f(x)

, but

h(c) = f(c)

, and since

f

is discontinuous, these likely won't be equal. Choice C creates the same problem as B since

f(c) \neq \lim_{x \to c} f(x)

for removable discontinuities. Remember: multiplying by zero "kills" discontinuities. When you see products involving discontinuous functions, consider whether one factor being zero can force continuity.

Question 7

Consider the piecewise function $f(x) = \begin{cases} \frac{x-1}{1-\sqrt{x}} & x \in [0, 1) \\ a & x=1 \\ \frac{\sin(\pi x)}{x-1} & x > 1 \end{cases}$ Which statement is true regarding the continuity of $f(x)$ at $x=1$ ?

The function is continuous at $x=1$ if $a=-2$ .
The function is continuous at $x=1$ if $a=-\pi$ .
The function has a jump discontinuity at $x=1$ for any value of $a$ . (correct answer)
The function has a removable discontinuity at $x=1$ if $a$ is the average of $-2$ and $-\pi$ .

Explanation: For continuity at

x=1

, the left-hand limit, the right-hand limit, and the function value must all be equal. Left-hand limit:

\lim_{x \to 1^-} \frac{x-1}{1-\sqrt{x}} = \lim_{x \to 1^-} \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{-(\sqrt{x}-1)} = \lim_{x \to 1^-} -(\sqrt{x}+1) = -2

. Right-hand limit:

\lim_{x \to 1^+} \frac{\sin(\pi x)}{x-1}

. Let

u=x-1

, so

x=u+1

. As

x \to 1^+

u \to 0^+

. The limit becomes

\lim_{u \to 0^+} \frac{\sin(\pi(u+1))}{u} = \lim_{u \to 0^+} \frac{\sin(\pi u + \pi)}{u} = \lim_{u \to 0^+} \frac{-\sin(\pi u)}{u} = \lim_{u \to 0^+} \frac{-\sin(\pi u)}{\pi u} \cdot \pi = -1 \cdot \pi = -\pi

. Since the left-hand limit (

-2

) and the right-hand limit (

-\pi

) are not equal, the function has a non-removable jump discontinuity at

x=1

, regardless of the value of

a

Question 8

A function $f(x) = \frac{P(x)}{Q(x)}$ , where $P(x)$ and $Q(x)$ are polynomials, has a removable discontinuity at $x=a$ and a vertical asymptote at $x=b$ . Which of the following statements must be true?

$P(a)=0$ , $Q(a)=0$ , and the multiplicity of the root $x=a$ in $P(x)$ is equal to its multiplicity in $Q(x)$ .
$P(a)=0$ , $Q(a)=0$ , and the multiplicity of the root $x=a$ in $P(x)$ is greater than or equal to its multiplicity in $Q(x)$ ; also $Q(b)=0$ . (correct answer)
$P(b) \neq 0$ , $Q(b)=0$ , and $P(a)=0$ .
$Q(a)=0$ , $Q(b)=0$ , and $P(b)=0$ .

Explanation: When analyzing discontinuities in rational functions, you need to understand how zeros in the numerator and denominator create different behaviors. The key is examining where each polynomial equals zero and comparing the multiplicities of shared roots. For a removable discontinuity at

x = a

, both

P(a) = 0

and

Q(a) = 0

must be true - otherwise you'd have either a continuous point or a vertical asymptote. The discontinuity is removable only when the factor

(x-a)

appears in both numerator and denominator, and the multiplicity in the numerator is at least as large as in the denominator. This allows you to "cancel out" the problematic factor. For a vertical asymptote at

x = b

, you need

Q(b) = 0

while

P(b) \neq 0

. If

P(b) = 0

as well, you'd have another removable discontinuity instead of an asymptote. Answer B correctly captures both conditions:

P(a) = 0

and

Q(a) = 0

with the numerator's multiplicity ≥ denominator's multiplicity at

x = a

, plus

Q(b) = 0

for the vertical asymptote. Answer A is too restrictive - equal multiplicities work, but greater multiplicity in the numerator also creates removable discontinuities. Answer C incorrectly states

P(a) = 0

without requiring

Q(a) = 0

, and doesn't address the asymptote condition. Answer D wrongly requires

P(b) = 0

, which would eliminate the vertical asymptote. Remember: removable discontinuities require shared zeros with sufficient multiplicity in the numerator, while vertical asymptotes need zeros only in the denominator.

Question 9

Let the function $f(x) = \frac{x^2 - 7x + 12}{x^2 - x - 6}$ . Let $g(x)$ be a new function that is continuous for all real numbers and is identical to $f(x)$ except at $x=3$ . What is the value of $g(3)$ ?

$-\frac{1}{5}$ (correct answer)
$0$
$\frac{1}{5}$
$\frac{7}{5}$

Explanation: To find the value that makes the function continuous, we must evaluate the limit as

x

approaches 3. First, factor the numerator and denominator:

f(x) = \frac{(x-3)(x-4)}{(x-3)(x+2)}

. For

x \neq 3

, we can cancel the

(x-3)

terms, leaving

\frac{x-4}{x+2}

. The limit is then

\lim_{x \to 3} \frac{x-4}{x+2} = \frac{3-4}{3+2} = -\frac{1}{5}

. Therefore, defining

g(3) = -1/5

will make the function continuous.

Question 10

Which of the following statements accurately describes the function $f(x) = \frac{x^2 - 9}{|x+3|}$ at $x=-3$ ?

It has a removable discontinuity, and the continuous extension has a value of 6 at $x=-3$ .
It has a removable discontinuity, and the continuous extension has a value of -6 at $x=-3$ .
It has a non-removable jump discontinuity because the left- and right-hand limits are not equal. (correct answer)
It has a non-removable infinite discontinuity.

Explanation: To analyze the function at

x=-3

, we must evaluate the one-sided limits. For

x \to -3^+

x+3 > 0

, so

|x+3| = x+3

. The limit is

\lim_{x \to -3^+} \frac{(x-3)(x+3)}{x+3} = \lim_{x \to -3^+} (x-3) = -6

. For

x \to -3^-

x+3 < 0

, so

|x+3| = -(x+3)

. The limit is

\lim_{x \to -3^-} \frac{(x-3)(x+3)}{-(x+3)} = \lim_{x \to -3^-} -(x-3) = -(-3-3) = 6

. Since the left-hand limit (6) does not equal the right-hand limit (-6), the function has a non-removable jump discontinuity at

x=-3

Question 11

The function $f(x) = \frac{ax^2 + bx + 4}{x-2}$ has a removable discontinuity at $x=2$ , and the graph of the continuous extension of $f(x)$ passes through the point $(2, 10)$ . What is the value of $a+b$ ?

$-8$ (correct answer)
$-2$
$6$
$10$

Explanation: For the discontinuity at

x=2

to be removable, the numerator must be zero at

x=2

. So,

a(2)^2 + b(2) + 4 = 0

, which simplifies to

4a + 2b = -4

, or

2a + b = -2

. The value of the continuous extension at

x=2

is the limit, which is given as 10. Using L'Hôpital's Rule on the

\frac{0}{0}

form:

\lim_{x \to 2} \frac{2ax + b}{1} = 10

. Substituting

x=2

gives

4a + b = 10

. We now have a system of two linear equations: (1)

2a + b = -2

and (2)

4a + b = 10

. Subtracting (1) from (2) gives

2a = 12

, so

a=6

. Substituting

a=6

into (1) gives

2(6) + b = -2

, so

12 + b = -2

, which means

b = -14

. The value of

a+b

6 + (-14) = -8

Question 12

Let $f(x) = p(x)/q(x)$ be a rational function where $p(x)$ and $q(x)$ are polynomials. If $f$ has a removable discontinuity at $x=c$ , what can be concluded about the multiplicities of the root $x=c$ for $p(x)$ and $q(x)$ ? Let $m$ be the multiplicity of the root $x=c$ in $p(x)$ and $n$ be the multiplicity in $q(x)$ .

$n > m \geq 0$
$m > n \geq 1$
$m = n \geq 1$
$m \geq n \geq 1$ (correct answer)

Explanation: For

f(x)

to have a discontinuity at

x=c

q(c)

must be 0, which means the multiplicity

n

of the root

c

q(x)

must be at least 1 (

n \geq 1

). The function can be written as

f(x) = \frac{(x-c)^m P(x)}{(x-c)^n Q(x)}

where

P(c), Q(c) \neq 0

. The limit as

x \to c

is determined by the term

(x-c)^{m-n}

. For a removable discontinuity, this limit must be finite. This occurs if the exponent

m-n

is greater than or equal to zero, i.e.,

m \geq n

. If

m>n

, the limit is 0. If

m=n

, the limit is

P(c)/Q(c)

. Both are finite limits, corresponding to removable discontinuities. Therefore, the condition is

m \geq n \geq 1

Question 13

Let $f(x) = \frac{(x-a)^3}{(x^2-a^2)(x-a)}$ for some non-zero constant $a$ . Which of the following statements best describes the discontinuity at $x=a$ ?

It is a removable discontinuity, and the continuous extension has a value of 0. (correct answer)
It is a removable discontinuity, and the continuous extension has a value of $\frac{a}{2}$ .
It is a non-removable infinite discontinuity.
The function is continuous at $x=a$ .

Explanation: First, simplify the expression for

f(x)

. The denominator is

(x-a)(x+a)(x-a) = (x-a)^2(x+a)

. So,

f(x) = \frac{(x-a)^3}{(x-a)^2(x+a)}

. For

x \neq a

, we can cancel

(x-a)^2

from the numerator and denominator, which gives

f(x) = \frac{x-a}{x+a}

. To determine the nature of the discontinuity at

x=a

, we evaluate the limit:

\lim_{x \to a} \frac{x-a}{x+a} = \frac{a-a}{a+a} = \frac{0}{2a} = 0

. Since

a

is non-zero, this limit is 0. Because the limit exists and is finite, the discontinuity is removable, and the value of the continuous extension at

x=a

is 0.

Question 14

Let $f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x}$ . To remove the discontinuity at $x=0$ , what value should be assigned to $f(0)$ ?

$0$
$\frac{1}{4}$
$4$
$\frac{1}{2}$ (correct answer)

Explanation: When you see a function that's undefined at a point but appears to have a pattern, you're dealing with a removable discontinuity. The key is finding the limit as x approaches that point. Since

f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x}

gives us

\frac{0}{0}

when we substitute

x = 0

, we need to use algebraic manipulation to find

\lim_{x \to 0} f(x)

. The classic technique here is rationalizing by multiplying by the conjugate:

f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x} \cdot \frac{\sqrt{4+x} + \sqrt{4-x}}{\sqrt{4+x} + \sqrt{4-x}}

Using the difference of squares formula

(a-b)(a+b) = a^2 - b^2

f(x) = \frac{(4+x) - (4-x)}{x(\sqrt{4+x} + \sqrt{4-x})} = \frac{2x}{x(\sqrt{4+x} + \sqrt{4-x})} = \frac{2}{\sqrt{4+x} + \sqrt{4-x}}

Now we can evaluate the limit:

\lim_{x \to 0} \frac{2}{\sqrt{4+0} + \sqrt{4-0}} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}

Answer choice A (

0

) would result from incorrectly thinking the numerator approaching 0 makes the whole limit 0. Choice B (

\frac{1}{4}

) might come from errors in the algebraic manipulation. Choice C (

4

) could result from confusing the coefficient in the square roots with the final answer. Remember: when you see an indeterminate form involving square roots, rationalization is usually your best strategy to reveal the true limit.

Question 15

Let $f(x) = \frac{x^2 - 1}{x-1}$ . The function $g(x) = \cos(\pi \cdot f(x))$ has a removable discontinuity at $x=1$ . What value should be assigned to $g(1)$ to make $g$ continuous at that point?

$-1$
$0$
$1$ (correct answer)
$2$

Explanation: First, find the limit of the inner function,

f(x)

, as

x \to 1

\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2

. Since the cosine function is continuous everywhere, we can find the limit of the composite function by passing the limit inside:

\lim_{x \to 1} g(x) = \cos(\pi \cdot \lim_{x \to 1} f(x)) = \cos(\pi \cdot 2) = \cos(2\pi) = 1

. To make

g(x)

continuous at

x=1

, we must define

g(1)

to be this limit.

Question 16

The function $f(x) = \frac{\sin(3x)}{x^2 + x}$ has a removable discontinuity at $x=0$ . If a new function $g(x)$ is defined such that it is continuous at $x=0$ and agrees with $f(x)$ for all $x \neq 0$ , what is the value of $g(0)$ ?

$0$
$1$
$3$ (correct answer)
The discontinuity is non-removable.

Explanation: To remove the discontinuity, we need to find the limit of

f(x)

x \to 0

. We can rewrite the function as

f(x) = \frac{\sin(3x)}{x(x+1)} = \frac{\sin(3x)}{x} \cdot \frac{1}{x+1}

. Using the known limit

\lim_{u \to 0} \frac{\sin(au)}{u} = a

, we have

\lim_{x \to 0} \frac{\sin(3x)}{x} = 3

. The limit of the second part is

\lim_{x \to 0} \frac{1}{x+1} = 1

. Therefore,

\lim_{x \to 0} f(x) = 3 \cdot 1 = 3

. Setting

g(0) = 3

makes the function continuous.

Question 17

Consider the function $f(x) = \frac{x^2 + kx - 10}{x-2}$ . If the function has a removable discontinuity at $x=2$ , what value must $f(2)$ be defined as to make the function continuous at that point?

3
5
7 (correct answer)
-5

Explanation: For

f(x)

to have a removable discontinuity at

x=2

, the numerator must be zero when

x=2

. This allows for the

(x-2)

factor in the denominator to be cancelled. So, we set the numerator to zero:

(2)^2 + k(2) - 10 = 0

, which gives

4 + 2k - 10 = 0

, or

2k=6

, so

k=3

. Now, we find the limit of the function as

x

approaches 2:

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = \lim_{x \to 2} (x+5) = 2+5=7

. To remove the discontinuity, we must define

f(2)

to be the value of the limit, which is 7.

Question 18

What value should be assigned to $f(0)$ to make the function $f(x) = \frac{1 - \cos^2(3x)}{x \sin(x)}$ continuous at $x=0$ ?

3
1
9 (correct answer)
0

Explanation: To find the value that removes the discontinuity at

x=0

, we need to evaluate the limit of

f(x)

x \to 0

. First, use the Pythagorean identity

1 - \cos^2(\theta) = \sin^2(\theta)

f(x) = \frac{\sin^2(3x)}{x \sin(x)}

. Now, we evaluate the limit:

\lim_{x \to 0} \frac{\sin^2(3x)}{x \sin(x)} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{\sin(3x)}{x} \cdot \frac{3x}{x} \cdot \frac{x}{\sin(x)}

is one way. A more direct way is:

\lim_{x \to 0} \frac{\sin^2(3x)}{(3x)^2} \cdot \frac{(3x)^2}{x \sin(x)} = \lim_{x \to 0} \left(\frac{\sin(3x)}{3x}\right)^2 \cdot \frac{9x^2}{x \sin(x)}

. Since

\lim_{u \to 0} \frac{\sin u}{u} = 1

, the first part of the product goes to

1^2=1

. The limit becomes:

\lim_{x \to 0} \frac{9x^2}{x \sin(x)} = \lim_{x \to 0} \frac{9x}{\sin(x)} = 9 \cdot \lim_{x \to 0} \frac{x}{\sin(x)} = 9 \cdot 1 = 9

. Therefore, to make the function continuous at

x=0

, we must define

f(0)=9

Question 19

The graph of the function $f(x) = \frac{\sqrt{x+a}-b}{x-4}$ has a removable discontinuity (a hole) at the point $(4, \frac{1}{6})$ . What are the values of the constants $a$ and $b$ ?

$a=5$ and $b=3$ (correct answer)
$a=32$ and $b=6$
$a=5$ and $b=-3$
$a=5$ and $b=1/6$

Explanation: For a removable discontinuity at

x=4

, the limit as

x \to 4

must exist, which requires the function to have a

0/0

indeterminate form. Thus, the numerator must be zero at

x=4

\sqrt{4+a}-b = 0

, which implies

b=\sqrt{4+a}

. The y-coordinate of the hole is the value of the limit. So,

\lim_{x \to 4} f(x) = \frac{1}{6}

. We evaluate the limit using the conjugate:

\lim_{x \to 4} \frac{\sqrt{x+a}-b}{x-4} = \lim_{x \to 4} \frac{\sqrt{x+a}-\sqrt{4+a}}{x-4} \cdot \frac{\sqrt{x+a}+\sqrt{4+a}}{\sqrt{x+a}+\sqrt{4+a}} = \lim_{x \to 4} \frac{(x+a)-(4+a)}{(x-4)(\sqrt{x+a}+\sqrt{4+a})} = \lim_{x \to 4} \frac{1}{\sqrt{x+a}+\sqrt{4+a}} = \frac{1}{2\sqrt{4+a}}

. We set this equal to

\frac{1}{6}

\frac{1}{2\sqrt{4+a}} = \frac{1}{6} \implies 2\sqrt{4+a}=6 \implies \sqrt{4+a}=3 \implies 4+a=9 \implies a=5

. Finally, using

b=\sqrt{4+a}

, we get

b=\sqrt{4+5}=\sqrt{9}=3

Question 20

The function $f(x) = \frac{\sin(2x)}{x^2+x}$ has discontinuities at $x=0$ and $x=-1$ . Which statement correctly describes how these discontinuities can be treated?

Only the discontinuity at $x=0$ is removable; it is removed by defining $f(0)=2$ . (correct answer)
Both discontinuities are removable; they are removed by defining $f(0)=2$ and $f(-1)=-\sin(2)$ .
Only the discontinuity at $x=0$ is removable; it is removed by defining $f(0)=1$ .
Only the discontinuity at $x=-1$ is removable; it is removed by defining $f(-1)=0$ .

Explanation: We analyze the limits at both points. At

x=0

\lim_{x \to 0} \frac{\sin(2x)}{x(x+1)} = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \frac{2}{x+1} = 1 \cdot \frac{2}{0+1} = 2

. Since the limit exists, the discontinuity at

x=0

is removable. It can be removed by defining

f(0)=2

. At

x=-1

\lim_{x \to -1} \frac{\sin(2x)}{x(x+1)}

. The numerator approaches

\sin(-2)

, which is a non-zero number. The denominator approaches 0. This results in an infinite discontinuity (a vertical asymptote), which is not removable. Therefore, only the discontinuity at

x=0

is removable.