Limits - Mathematics Questions and Answers

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Limits”.

1. What is the value of \(\lim\limits_{y \rightarrow 2} \frac{y^2-4}{y-2}\)?
a) 2
b) 4
c) 1
d) 0
View Answer

Answer: b
Explanation: y² – 4 = (y – 2)(y + 2)
Therefore the fraction becomes, (y + 2)
As y tends to 2, the fraction becomes 4

2. What is the value of \(\lim\limits_{y \rightarrow \infty} \frac{2}{y}\)?
a) 0
b) 1
c) 2
d) Infinity
View Answer

Answer: a
Explanation: Any number divided by infinity gives us 0.
Here, since the number 2 is divided by y, as y approaches infinity, we get 0

3. What is the value of \(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\)?
a) 0
b) 2
c) 8
d) 6
View Answer

Answer: d
Explanation: The denominator becomes 0, as x approaches 4.
\(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\) Here, if we factorize the numerator we get:
\(\lim\limits_{x \rightarrow 4} \frac{(x – 4)(x + 2)}{x – 4}\)
We can now cancel out (x – 4) from both the numerator and denominator.
We get, \(\lim\limits_{x \rightarrow 4}\)(x + 2) = 6

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4. What is the value of \(\lim\limits_{x \rightarrow 3}\frac{x^2-9}{x–3}\)?
a) 0
b) 3
c) Infinity
d) 6
View Answer

Answer: d
Explanation: When x tends to 3, both the numerator and the denominator become 0 and it becomes of the form, \(\frac{0}{0}\).
Therefore, we use L’Hospital’s rule, which states the we differentiate the numerator and the denominator, until a definite answer is reached.
On differentiating once we get,
\(\lim\limits_{x \rightarrow 3}\frac{2x}{1}\)
Since, this not an indeterminate form now, we can substitute the value of x.
= 2 x 3
= 6

5. What is the value of \(\lim\limits_{x \rightarrow \infty}\frac{x^2-9}{x^2–3x+2}\)?
a) 1
b) 2
c) 0
d) Limit does not exist
View Answer

Answer: a
Explanation:
Since it is of the form \(\frac{\infty}{\infty}\), we use L’Hospital’s rule and differentiate the numerator and denominator
L = \(\lim\limits_{x \rightarrow \infty}\frac{x^2-9}{x^2–3x+2}\)
On differentiating once, we get \(\lim\limits_{x \rightarrow \infty}\frac{2x}{2x}\)
Which is equal to, \(\lim\limits_{x \rightarrow \infty}\) ⁡1 = 1.

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6. Which of the following limits does not yield 1?
a) \(\lim\limits_{x \rightarrow 0}\)⁡ 1
b) \(\lim\limits_{x \rightarrow \infty}\)x^-2 + x^-1 + 1
c) \(\lim\limits_{x \rightarrow \infty}\frac{1}{e^x}\) + 1
d) \(\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}\)
View Answer

Answer: d
Explanation: \(\lim\limits_{x \rightarrow 0}\)⁡ 1 = 1 (Since no x term is present)
When the denominator is infinity, the value of the fraction is 0, provided the numerator is not infinity.
\(\lim\limits_{x \rightarrow \infty}\)x^-2 + x^-1 + 1 = 0 + 0 + 1 = 1
\(\lim\limits_{x \rightarrow \infty}\frac{1}{e^x}\) + 1 = 1 ( e^-∞ = 0)
\(\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}\) (Use L’Hospital’s rule and differentiate the numerator and denominator until a rational form is obtained)
\(\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}\) = \(\lim\limits_{x \rightarrow \infty}\frac{3x^2+2x+32}{2x–3}\) = \(\lim\limits_{x \rightarrow \infty}\frac{3x}{2}\) = ∞

7. What is the value of \(\lim\limits_{y \rightarrow 4}\) f(y)? It is given that f(y) = y² + 6y (y ≥ 2) and f(y) = 0 (y < 2).
a) 40
b) 16
c) 0
d) 30
View Answer

Answer: a
Explanation: \(\lim\limits_{y \rightarrow 4}\)f(y) = y² + 6y
f(4) = 4² + 6(4)
f(4) = 16 + 24
f(4) = 40

8. What is the value of the limit f(x) = \(\frac{x^2+\sqrt {2x}}{x^2-4x}\) if x approaches infinity?
a) 0
b) 2
c) 1/2
d) 4
View Answer

Answer: a
Explanation: This is of the form \(\frac{\infty}{\infty}\), therefore we use L’Hospital’s rule and differentiate the numerator and denominator.
= \(\lim\limits_{x \rightarrow \infty}\frac{2x+\sqrt{2/x}}{2x–4}\)
= \(\lim\limits_{x \rightarrow \infty}⁡\)√2 x^-3/2
= 0

9. What is the value of the \(\lim\limits_{x \rightarrow 5}⁡\frac{32x+1}{x^2–5x}\)?
a) 6.2
b) 6.4
c) 6.3
d) 6.1
View Answer

Answer: b
Explanation: Use L’Hospital’s Rule, and differentiate the numerator and denominator.
\(\lim\limits_{x \rightarrow 5}\frac{⁡32}{2x–5}\)
= \(\frac{32}{5}\)
= 6.4

10. What is the value of the limit \(\lim\limits_{x \rightarrow 4}\frac{x^2-4-3x}{x-3}\)?
a) 0
b) 4
c) 1
d) Limit does not exist
View Answer

Answer: a
Explanation: \(\lim\limits_{x \rightarrow 4}\frac{x^2-4-3x}{x-3}\)
= \(\frac{4^2-4-3(4)}{4-3}\)
= \(\frac{0}{1}\)
= 0.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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