250+ TOP MCQs on Limits of Trigonometric Functions & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Limits of Trigonometric Functions”.

1. What is the value of (limlimits_{y rightarrow pi/2}frac{sin⁡ x}{x})?
a) (frac{2}{pi})
b) (frac{pi}{2})
c) 1
d) 0
Answer: a
Clarification: sin ⁡(frac{pi}{2}) = 1
(limlimits_{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 (limlimits_{y rightarrow 0}frac{sin⁡3y}{3y})?
a) 0
b) 1
c) 3
d) (frac{1}{3})
Answer: b
Clarification: We know that (limlimits_{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.
= (limlimits_{y rightarrow 0}frac{3, cos, 3y}{3})
= 1

3. What is the value of (limlimits_{x rightarrow 0}frac{x^2sec x}{sin⁡ x})?
a) 3
b) 2
c) 1
d) 0
Answer: d
Clarification: (limlimits_{x rightarrow 0}frac{x}{sin⁡ x})x (limlimits_{x rightarrow 0}⁡frac{x}{cos⁡ x})
= 1 x 0
= 0

4. What is the value of (limlimits_{x rightarrow 0}frac{x ,tanx}{cot, x})?
a) 0
b) 1
c) 2
d) (frac{1}{2})
Answer: a
Clarification: (limlimits_{x rightarrow 0}frac{x tanx}{cot x}) = (limlimits_{x rightarrow 0}frac{xfrac{sin⁡ x}{cos ⁡x}}{frac{cos⁡ x}{sin⁡ x}})
= (limlimits_{x rightarrow 0}) ⁡x
= 0

5. What is the value of (limlimits_{x rightarrow infty}frac{x sin⁡frac{2}{x}}{2})?
a) 1
b) 2
c) (frac{1}{2})
d) Limit does not exist
Answer: a
Clarification:
This is of the form (frac{0}{0}), so we use L’Hospital’s rule.
= (limlimits_{x rightarrow infty}frac{frac{-2}{x^2}cos⁡frac{2}{x}}{frac{-2}{x^2}})
= (limlimits_{x rightarrow infty})cos(frac{2}{x})
= 1

6. Which of the following limits does not yield 1?
a) (limlimits_{x rightarrow 0}frac{⁡sin x}{x})
b) (limlimits_{x rightarrow 0}frac{⁡tan x}{cot x})
c) (limlimits_{x rightarrow 0}(frac{1}{e^x}+cos⁡ x))
d) (limlimits_{x rightarrow 0}) x cosec x
Answer: c
Clarification: (limlimits_{x rightarrow 0}(frac{1}{e^x} + sin⁡ x) = frac{1}{e^0}) + cos (0)
= 1 + 1
= 2

7. What is the value of (limlimits_{y rightarrow 0})(32 x² cosec² ⁡4x)?
a) 1
b) 4
c) 2
d) 3
Answer: c
Clarification: The limit can be written as, (limlimits_{x rightarrow 0}frac{32x^2}{sin^2⁡4x})
= 2 x (limlimits_{x rightarrow 0}frac{4x}{sin 4x}) x (limlimits_{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}{2sqrt 2})
d) (frac{1}{2sqrt 2})
Answer: c
Clarification: This is of the form (frac{0}{0}), therefore we use L’Hospital’s rule and differentiate the numerator and denominator.
= (limlimits_{x rightarrow 0}frac{2sin⁡ ,x cos ,⁡x + cos ,⁡x sqrt 2}{2x – 4})
= (frac{0+sqrt 2}{-4})
= (frac{-1}{2sqrt 2})

9. What is the value of the (limlimits_{x rightarrow frac{3pi}{2}}frac{cos⁡ x sin⁡ x}{sin⁡2x})?
a) (frac{-1}{2})
b) (frac{1}{2})
c) (frac{1}{4})
d) (frac{-1}{4})
Answer: b
Clarification: (limlimits_{x rightarrow frac{3pi}{2}}frac{cos⁡ x sin⁡ x}{sin⁡2x}) =(limlimits_{x rightarrow frac{3pi}{2}}frac{cos⁡ x sin⁡ x}{2 cos x sin⁡ x})
= (frac{1}{2})

10. What is the value of the limit (limlimits_{x rightarrow frac{pi}{2}}frac{sin^2⁡x-1}{cos ⁡x})?
a) 0
b) 4
c) 1
d) Limit does not exist
Answer: a
Clarification: (limlimits_{x rightarrow frac{pi}{2}}frac{sin^2⁡x-1}{cos ⁡x}) = (limlimits_{x rightarrow frac{pi}{2}}frac{-cos^2 x}{cos ⁡x})
=(limlimits_{x rightarrow frac{pi}{2}}) -cosx
= 0

11. What is the value of the limit (limlimits_{x rightarrow 0}frac{sin^2⁡x}{x^2})?
a) 2
b) 1
c) Limit does not exist
d) 4
Answer: b
Clarification: (limlimits_{x rightarrow 0}frac{sin^2⁡x}{x^2}) =
= ((limlimits_{x rightarrow 0}frac{sin ⁡x}{x}) x (limlimits_{x rightarrow 0}frac{sinx}{x}))
We apply L’Hospital’s rule and differentiate numerator and denominator.
= ((limlimits_{x rightarrow 0}frac{cos x}{1}) x (limlimits_{x rightarrow 0}frac{cos x}{1}))
= 1

12. What is the value of (limlimits_{x rightarrow 0}frac{e^x(sin^2⁡ x)}{x^3})?
a) 2
b) 3
c) 1
d) 0
Answer: c
Clarification: (limlimits_{x rightarrow 0}frac{sin^2⁡ x}{x^2}) x (limlimits_{x rightarrow 0}frac{e^x}{x})
We apply L’Hospital’s rule and differentiate numerator and denominator.
= 1 x (limlimits_{x rightarrow 0}frac{e^x}{1})
= 1