Mathematics Questions and Answers – Limits of Trigonometric Functions

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This set of Mathematics 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
\(\lim\limits_{y \rightarrow \pi/2}\frac{sin⁡x}{x} = \frac{sin⁡\frac{π}{2}}{\frac{\pi}{2}}\)
= \(\frac{1}{\frac{\pi}{2}}\)
= \(\frac{2}{\pi}\)
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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
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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
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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: \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\) =\(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{2 cos x sin⁡ x}\)
= \(\frac{1}{2}\)
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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

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter