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

Mathematics Multiple Choice Questions on “Limits”.

1. What is the value of (limlimits_{y rightarrow 2} frac{y^2-4}{y-2})?
a) 2
b) 4
c) 1
d) 0
Answer: b
Clarification: 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 (limlimits_{y rightarrow infty} frac{2}{y})?
a) 0
b) 1
c) 2
d) Infinity
Answer: a
Clarification: 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 (limlimits_{x rightarrow 4} frac{x^2-2x-8}{x-4})?
a) 0
b) 2
c) 8
d) 6
Answer: d
Clarification: The denominator becomes 0, as x approaches 4.
(limlimits_{x rightarrow 4} frac{x^2-2x-8}{x-4}) Here, if we factorize the numerator we get:
(limlimits_{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, (limlimits_{x rightarrow 4})(x + 2) = 6

4. What is the value of (limlimits_{x rightarrow 3}frac{x^2-9}{x–3})?
a) 0
b) 3
c) Infinity
d) 6
Answer: b
Clarification: 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,
(limlimits_{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 (limlimits_{x rightarrow infty}frac{x^2-9}{x^2–3x+2})?
a) 1
b) 2
c) 0
d) Limit does not exist
Answer: a
Clarification:
Since it is of the form (frac{infty}{infty}), we use L’Hospital’s rule and differentiate the numerator and denominator
L = (limlimits_{x rightarrow infty}frac{x^2-9}{x^2–3x+2})
On differentiating once, we get (limlimits_{x rightarrow infty}frac{2x}{2x})
Which is equal to, (limlimits_{x rightarrow infty}) ⁡1 = 1.

6. Which of the following limits does not yield 1?
a) (limlimits_{x rightarrow 0})⁡ 1
b) (limlimits_{x rightarrow infty})x^-2 + x^-1 + 1
c) (limlimits_{x rightarrow infty}frac{1}{e^x}) + 1
d) (limlimits_{x rightarrow infty}frac{x^3+x^2+32x+1}{x^2–3x+2})
Answer: d
Clarification: (limlimits_{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.
(limlimits_{x rightarrow infty})x^-2 + x^-1 + 1 = 0 + 0 + 1 = 1
(limlimits_{x rightarrow infty}frac{1}{e^x}) + 1 = 1 ( e^-∞ = 0)
(limlimits_{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)
(limlimits_{x rightarrow infty}frac{x^3+x^2+32x+1}{x^2–3x+2}) = (limlimits_{x rightarrow infty}frac{3x^2+2x+32}{2x–3}) = (limlimits_{x rightarrow infty}frac{3x}{2}) = ∞

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

9. What is the value of the (limlimits_{x rightarrow 5}⁡frac{32x+1}{x^2–5x})?
a) 6.2
b) 6.4
c) 6.3
d) 6.1
Answer: b
Clarification: Use L’Hospital’s Rule, and differentiate the numerator and denominator.
(limlimits_{x rightarrow 5}frac{⁡32}{2x–5})
= (frac{32}{5})
= 6.4

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