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📝 Limits of rational functions (18 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 18 questions available

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What is Limits of rational functions?

Definition:
The limit of a rational function $f(x) = \frac{P(x)}{Q(x)}$ , where $P$ and $Q$ are polynomials, as $x \to a$ is evaluated by direct substitution if $Q(a) \neq 0$ , yielding $\frac{P(a)}{Q(a)}$ . If $Q(a) = 0$ but $P(a) \neq 0$ , the limit is infinite or does not exist; if both are zero, factorization or cancellation is needed to resolve the indeterminate form.

Example:
Evaluate $\lim_{x \to 2} \frac{x^2-4}{x-2}$ .
Solution: Direct substitution gives $0/0$ , so factor: $\frac{(x-2)(x+2)}{x-2} = x+2$ , then $\lim_{x \to 2} (x+2) = 4$ .

Reason:
Understanding rational function limits is essential because they appear frequently in calculus, and handling cases where the denominator vanishes teaches algebraic manipulation and introduces the concept of removable discontinuities.

Easy

Medium

Hard

📝 All Limits of rational functions MCQs

Q1. For $f(x)=\dfrac{2-x}{(x-4)(x+2)}$ , what is the sign of the limit as $x\to 4^{+}$ ?

A.#NAME?

B.#NAME? ✅

C.0

D.Does not exist

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: As $x$ approaches 4 from the right, the factor $2-x$ is negative, $(x-4)$ is positive, and $(x+2)$ is positive, making the denominator positive. A negative numerator over a positive denominator yields a negative quantity that grows without bound, so the limit is $-\infty$ .

Q2. Which of the following best defines a vertical asymptote of a rational function?

A.A point where function is undefined and limit is finite

B.A line

x = a

where the function grows without bound as

x

approaches

a

✅

C.A line

y = b

where function approaches

b

x\to\infty

D.A point where derivative does not exist

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A vertical asymptote occurs at a value $x=a$ where the denominator approaches zero while the numerator does not, causing the function values to increase or decrease without bound as $x$ gets arbitrarily close to $a$ from either side.

Q3. Compare the one‑sided limits of $p(x)=\frac{1}{x-2}$ and $q(x)=\frac{1}{(x-2)^{2}}$ as $x\to 2^{+}$ . Which statement is true?

A.Both limits equal

+\infty

✅

B.Limit of

p(x)

+\infty

while limit of

q(x)

-\infty

C.Limit of

p(x)

-\infty

while limit of

q(x)

+\infty

D.Neither limit exists

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When $x$ approaches 2 from the right, $x-2$ is a small positive number. Its reciprocal is a large positive number, so $p(x)\to+\infty$ . Squaring the denominator makes it positive regardless of sign, so $q(x)$ also tends to $+\infty$ . Hence both limits are $+\infty$ .

Q4. Evaluate $\displaystyle\lim_{x\to 3}\frac{x^{2}-6x+9}{x-3}$ by simplifying the expression.

A.0 ✅

B.3

C.6

D.Limit does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The numerator factors as $(x-3)^{2}$ . Cancelling one factor of $(x-3)$ with the denominator leaves $x-3$ . As $x$ approaches 3, this remaining factor approaches 0, so the limit is 0.

Q5. Given $g(x)=\frac{2-x}{(x-4)(x+2)}$ , determine the sign of $\displaystyle\lim_{x\to 4^{-}} g(x)$ .

A.#NAME? ✅

B.#NAME?

C.0

D.Does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $x$ just left of 4, $2-x$ is negative, $(x-4)$ is negative, and $(x+2)$ is positive. The product of the two denominator factors is negative, so a negative numerator divided by a negative denominator gives a positive value that grows without bound, yielding $+\infty$ .

Q6. According to Theorem 1.2.2(d), what can be concluded if $\displaystyle\lim_{x\to a} q(x)=0$ while $\displaystyle\lim_{x\to a} p(x)\neq0$ for a rational function $\frac{p(x)}{q(x)}$ ?

A.The limit exists and equals 0

B.The limit does not exist and is infinite ✅

C.The limit equals the numerator limit

D.The limit equals the denominator limit

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the denominator approaches zero while the numerator approaches a non‑zero finite number, the quotient grows without bound. Depending on the sign of the approaching values, the limit is either $+\infty$ or $-\infty$ ; in any case, the limit does not exist as a finite real number.

Q7. Find $\displaystyle\lim_{x\to\infty}\frac{5x^{3}+4x-3}{2x^{3}+x^{2}+7}$ .

A.05-Feb ✅

B.02-May

C.1

D.Limit does not exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For large $x$ , the highest‑degree terms dominate. Dividing numerator and denominator by $x^{3}$ gives $\frac{5+4/x^{2}-3/x^{3}}{2+1/x+7/x^{3}}$ . As $x\to\infty$ , the lower‑order terms vanish, leaving the ratio of leading coefficients $5/2$ .

Q8. Let $h(x)=\frac{x^{2}-4x+3}{x^{2}-9}$ . After canceling common factors, which function gives the same limit as $x\to3$ ?

\frac{x-1}{x+3}

✅

\frac{x+1}{x-3}

\frac{x-3}{x+3}

\frac{x-1}{x-3}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Both numerator and denominator have a factor $(x-3)$ . Cancelling this factor yields $\frac{x-1}{x+3}$ . The limit of the original function as $x\to3$ equals the limit of the simplified function, because the removed factor is zero only at the point of evaluation, not in the limiting process.

Q9. For $k(x)=\frac{x^{3}-27}{(x-3)^{2}}$ , does $\displaystyle\lim_{x\to3}k(x)$ exist? Choose the correct description.

A.Limit is finite and equals 0

B.Limit is

+\infty

C.Limit is

-\infty

D.Limit does not exist because one‑sided limits differ ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Factor the numerator as $(x-3)(x^{2}+3x+9)$ and cancel one $(x-3)$ , leaving $\frac{x^{2}+3x+9}{x-3}$ . As $x$ approaches 3 from the right, the denominator is positive, giving $+\infty$ ; from the left it is negative, giving $-\infty$ . Since the one‑sided limits differ, the two‑sided limit does not exist.

Q10. Determine whether the two‑sided limit exists for $f(x)=\frac{2-x}{(x-4)(x+2)}$ at $x=4$ .

A.Exists and equals

-\infty

B.Exists and equals

+\infty

C.Exists and equals 0

D.Does not exist ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: From the right of 4 the expression is negative and unbounded, while from the left it is positive and unbounded. Because the one‑sided limits approach opposite infinities, the overall limit at $x=4$ does not exist.

Q11. Why does canceling a common factor $(x-a)$ from numerator and denominator not change the limit of a rational function as $x\to a$ ?

A.Because the factor equals zero at

x=a

B.Because limits depend only on values arbitrarily close to

a

, not at

a

✅

C.Because the canceled factor dominates the behavior

D.Because the function becomes constant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A limit concerns the behavior of a function as the variable approaches a point, not the function's value at that point. Removing a factor that is zero at the point eliminates a removable discontinuity but leaves the surrounding behavior unchanged, so the limit remains the same.

Q12. What is $\displaystyle\lim_{x\to5^{+}}\frac{1}{(x-5)^{2}}$ ?

A.#NAME? ✅

B.#NAME?

C.0

D.Does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The denominator $(x-5)^{2}$ is always positive. As $x$ approaches 5 from the right, the denominator becomes an increasingly small positive number, causing the fraction to grow without bound positively, i.e., $+\infty$ .

Q13. Compute $\displaystyle\lim_{x\to2}\frac{x^{2}-4}{x-2}$ .

A.2

B.4 ✅

C.0

D.Does not exist

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Factor the numerator as $(x-2)(x+2)$ . Cancel the common factor $(x-2)$ to obtain $x+2$ . Substituting $x=2$ gives $2+2=4$ . Hence the limit equals 4.

Q14. If $\displaystyle\lim_{x\to a}p(x)=5$ and $\displaystyle\lim_{x\to a}q(x)=3$ with $q(a)\neq0$ , what is $\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}$ ?

A.5

B.3

C.15-Sept

D.05-Mar ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The limit of a quotient equals the quotient of the limits provided the denominator limit is non‑zero. Here the quotient of the limits is $\frac{5}{3}$ .

Q15. Using the sign chart for $r(x)=\frac{2-x}{(x-4)(x+2)}$ , which interval yields a positive value of the function?

(- \infty,-2)

✅

(-2,4)

(4,\infty)

D.All intervals

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $x<-2$ , $2-x$ is positive, $(x-4)$ is negative, and $(x+2)$ is negative, making the denominator positive (negative × negative). Positive numerator divided by positive denominator gives a positive function value on that interval.

Q16. Evaluate $\displaystyle\lim_{x\to3}\frac{x^{3}-27}{x-3}$ .

A.27 ✅

B.9

C.0

D.Does not exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Factor the numerator as $(x-3)(x^{2}+3x+9)$ and cancel $(x-3)$ . The remaining expression is $x^{2}+3x+9$ . Substituting $x=3$ yields $9+9+9=27$ .

Q17. Find $\displaystyle\lim_{x\to2}\frac{x^{4}-16}{x^{2}-4}$ by simplifying the rational expression.

A.8 ✅

B.4

C.0

D.Does not exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Factor numerator as $(x^{2}-4)(x^{2}+4)$ and denominator as $(x^{2}-4)$ . Cancel the common factor to obtain $x^{2}+4$ . Substituting $x=2$ gives $4+4=8$ .

Q18. The discontinuity of $f(x)=\frac{x^{2}-4}{x-2}$ at $x=2$ is of which type?

A.Removable ✅

B.Jump

C.Infinite

D.Oscillatory

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Both numerator and denominator share the factor $(x-2)$ . After canceling, the function simplifies to $x+2$ , which is defined at $x=2$ . The original undefined point can be filled in

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