Limits of Trigonometric Functions - Class 11 Maths MCQ

This set of Class 11 Maths Chapter 13 Multiple Choice Questions & Answers (MCQs) focuses on “Limits of Trigonometric Functions”.

1. What is the value of \(\lim\limits_{y \rightarrow \pi/2}\frac{sin⁡ x}{x}\)?
a) \(\frac{2}{\pi}\)
b) \(\frac{\pi}{2}\)
c) 1
d) 0
View Answer

Answer: a
Explanation: sin ⁡\(\frac{\pi}{2}\) = 1
\(\lim\limits_{y \rightarrow \pi/2}\frac{sin⁡x}{x} = \frac{sin⁡\frac{π}{2}}{\frac{\pi}{2}}\)
= \(\frac{1}{\frac{\pi}{2}}\)
= \(\frac{2}{\pi}\)

2. What is the value of \(\lim\limits_{y \rightarrow 0}\frac{sin⁡3y}{3y}\)?
a) 0
b) 1
c) 3
d) \(\frac{1}{3}\)
View Answer

Answer: b
Explanation: We know that \(\lim\limits_{x \rightarrow 0}\frac{sin⁡x}{x}\) = 1.
Here x tends to 3y.
Also, since this is of the form \(\frac{0}{0}\), we use L’Hospital’s rule and differentiate the numerator and denominator separately.
= \(\lim\limits_{y \rightarrow 0}\frac{3\, cos\, 3y}{3}\)
= 1

3. What is the value of \(\lim\limits_{x \rightarrow 0}\frac{x^2sec x}{sin⁡ x}\)?
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: d
Explanation: \(\lim\limits_{x \rightarrow 0}\frac{x}{sin⁡ x}\)x \(\lim\limits_{x \rightarrow 0}⁡\frac{x}{cos⁡ x}\)
= 1 x 0
= 0

4. What is the value of \(\lim\limits_{x \rightarrow 0}\frac{x \,tanx}{cot\, x}\)?
a) 0
b) 1
c) 2
d) \(\frac{1}{2}\)
View Answer

Answer: a
Explanation: \(\lim\limits_{x \rightarrow 0}\frac{x tanx}{cot x}\) = \(\lim\limits_{x \rightarrow 0}\frac{x\frac{sin⁡ x}{cos ⁡x}}{\frac{cos⁡ x}{sin⁡ x}}\)
= \(\lim\limits_{x \rightarrow 0}\) ⁡x
= 0

5. What is the value of \(\lim\limits_{x \rightarrow \infty}\frac{x sin⁡\frac{2}{x}}{2}\)?
a) 1
b) 2
c) \(\frac{1}{2}\)
d) Limit does not exist
View Answer

Answer: a
Explanation:
This is of the form \(\frac{0}{0}\), so we use L’Hospital’s rule.
= \(\lim\limits_{x \rightarrow \infty}\frac{\frac{-2}{x^2}cos⁡\frac{2}{x}}{\frac{-2}{x^2}}\)
= \(\lim\limits_{x \rightarrow \infty}\)cos\(\frac{2}{x}\)
= 1

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6. Which of the following limits does not yield 1?
a) \(\lim\limits_{x \rightarrow 0}\frac{⁡sin x}{x}\)
b) \(\lim\limits_{x \rightarrow 0}\frac{⁡tan x}{cot x}\)
c) \(\lim\limits_{x \rightarrow 0}(\frac{1}{e^x}+cos⁡ x)\)
d) \(\lim\limits_{x \rightarrow 0}\) x cosec x
View Answer

Answer: c
Explanation: \(\lim\limits_{x \rightarrow 0}(\frac{1}{e^x} + sin⁡ x) = \frac{1}{e^0}\) + cos (0)
= 1 + 1
= 2

7. What is the value of \(\lim\limits_{y \rightarrow 0}\)(32 x² cosec² ⁡4x)?
a) 1
b) 4
c) 2
d) 3
View Answer

Answer: c
Explanation: The limit can be written as, \(\lim\limits_{x \rightarrow 0}\frac{32x^2}{sin^2⁡4x}\)
= 2 x \(\lim\limits_{x \rightarrow 0}\frac{4x}{sin 4x}\) x \(\lim\limits_{x \rightarrow 0}\frac{4x}{sin 4x}\)
= 2 x 1 x 1
= 2

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

Answer: c
Explanation: This is of the form \(\frac{0}{0}\), therefore we use L’Hospital’s rule and differentiate the numerator and denominator.
= \(\lim\limits_{x \rightarrow 0}\frac{2sin⁡ \,x cos \,⁡x + cos \,⁡x \sqrt 2}{2x – 4}\)
= \(\frac{0+\sqrt 2}{-4}\)
= \(\frac{-1}{2\sqrt 2}\)

9. What is the value of the \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\)?
a) \(\frac{-1}{2}\)
b) \(\frac{1}{2}\)
c) \(\frac{1}{4}\)
d) \(\frac{-1}{4}\)
View Answer

Answer: b
Explanation: \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\) =\(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{2 cos x sin⁡ x}\)
= \(\frac{1}{2}\)

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

Answer: a
Explanation: \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{sin^2⁡x-1}{cos ⁡x}\) = \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-cos^2 x}{cos ⁡x}\)
=\(\lim\limits_{x \rightarrow \frac{\pi}{2}}\) -cosx
= 0

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

Answer: b
Explanation: \(\lim\limits_{x \rightarrow 0}\frac{sin^2⁡x}{x^2}\) =
= (\(\lim\limits_{x \rightarrow 0}\frac{sin ⁡x}{x}\) x \(\lim\limits_{x \rightarrow 0}\frac{sinx}{x}\))
We apply L’Hospital’s rule and differentiate numerator and denominator.
= (\(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\) x \(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\))
= 1

12. What is the value of \(\lim\limits_{x \rightarrow 0}\frac{e^x(sin^2⁡ x)}{x^3}\)?
a) 2
b) 3
c) 1
d) 0
View Answer

Answer: c
Explanation: \(\lim\limits_{x \rightarrow 0}\frac{sin^2⁡ x}{x^2}\) x \(\lim\limits_{x \rightarrow 0}\frac{e^x}{x}\)
We apply L’Hospital’s rule and differentiate numerator and denominator.
= 1 x \(\lim\limits_{x \rightarrow 0}\frac{e^x}{1}\)
= 1

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