Class 11 Maths Chapter 13 Limits and Derivatives MCQ

This set of Class 11 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Limits and Derivatives”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

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

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.

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