$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}}$

exists and equals $\frac{1}{{2\sqrt 2 }}$

exists and equals $\frac{1}{{2\sqrt 2 (\sqrt 2 + 1)}}$

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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 930 questions

1 Medium

1 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

is equal to

View answer and explanation

Correct option: C

\frac{-1}{3}

Option C (

\frac{-1}{3}

) is correct because after rewriting the expression and applying L'Hôpital's rule multiple times, the limit evaluates to

\frac{-1}{3}

2 Medium

2 of 930

Analyzing +4 | -1

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

View answer and explanation

Correct option: C

\infty

Option C (∞) is correct because as x approaches pi/2, the expression tends towards infinity.

3 Easy

3 of 930

Analyzing +4 | -1

The value of

\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

View answer and explanation

Correct option: A

\frac{4}{3}

Option A (

\frac{4}{3}

) is correct because by applying trigonometric identities and L'Hopital's rule, the limit evaluates to

\frac{4}{3}

4 Medium

4 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because by rewriting the expression and applying L'Hopital's rule, the limit evaluates to 1.

5 Easy

5 of 930

Application +4 | -1

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

, then the value of

f^{\prime}(0)

is equal to

View answer and explanation

Correct option: C

Option C (13) is correct because it is the value of

f'(0)

found by logarithmic differentiation.

6 Medium

6 of 930

Application +4 | -1

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}

equals

View answer and explanation

Correct option: A

{e^a}

Option A (

{e^a}

) is correct because by rewriting the expression in the form of the exponential limit, the limit can be evaluated to

{e^a}

7 Medium

7 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow a}}\left[\left\{\left(\frac{\sqrt{a}+\sqrt{x}}{a^{1 / 4}-x^{1 / 4}}\right)^{-1}-\frac{2(a x)^{1 / 4}}{x^{3 / 4}-a^{1 / 4} x^{1 / 2}+a^{1 / 2} x^{1 / 4}-a^{3 / 4}}\right\}^{-1}\right.\left.-(\sqrt{2})^{\log _4 a}\right]^8

, is

View answer and explanation

Correct option: C

a^2

Option C (

a^2

) is correct because after simplification and taking the limit, the expression evaluates to

a^2

8 Medium

8 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\log _e \cot \left(\frac{\pi}{4}-K_1 x\right)}{\tan K_2 x}=1

, then

View answer and explanation

Correct option: B

2 K_1=K_2

The correct answer is B (

2 K_1=K_2

) because by applying logarithmic properties and limits, the relationship between

K_1

and

K_2

is found to be

2K_1 = K_2

9 Medium

9 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}(\cos x+\sin x)^{1 / x}

equals

View answer and explanation

Correct option: A

e

Option A (

e

) is correct because by taking the logarithm and applying L'Hopital's rule, the limit is found to be

e

10 Medium

10 of 930

Analyzing +4 | -1

Let

m

and

n

be two positive integers greater than 1. If

\lim \limits_{{\alpha \rightarrow 0}}\left(\frac{e^{\cos \left(\alpha^n\right)}-e}{\alpha^m}\right)=-\frac{e}{2}

, than the value of

\frac{m}{n}

, is

View answer and explanation

Correct option: A

Option A (2) is correct because the limit can be evaluated using L'Hopital's rule or Taylor series expansion, leading to the condition m = 2n, and thus m/n = 2.

11 Medium

11 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}} f(x)

, where

\frac{2 x-3}{x} < f(x) < \frac{2 x^2+5 x}{x^2}

, is-

View answer and explanation

Correct option: B

Option B (2) is correct because the limit is evaluated using the Squeeze Theorem (Sandwich Theorem).

12 Easy

12 of 930

Application +4 | -1

The value of

\lim\limits_{x \to \infty } \frac{{\log x}}{{{x^n}}},\;n > 0

View answer and explanation

Correct option: A

Option A (0) is correct because applying L'Hopital's rule to the limit results in 0.

13 Easy

13 of 930

Application +4 | -1

f(x)=\left\{\begin{array}{cl}\frac{\sin (1+[x])}{[x]} & \text {for }[x] \neq 0 \\ 0 & \text {for }[x]=0\end{array}\right.

where

[x]

denotes the greatest integer

\leq x

, then

\lim \limits_{{x \rightarrow 0-}} f(x)

equals

View answer and explanation

Correct option: B

The correct answer is B (0) because as x approaches 0 from the left, the greatest integer function [x] equals -1, and the limit is calculated accordingly.

14 Easy

14 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)

View answer and explanation

Correct option: B

equals 0

Option B (equals 0) is correct because the limit can be evaluated using the Squeeze Theorem and standard limits.

15 Medium

15 of 930

Analyzing +4 | -1

L=\lim \limits_{{x \rightarrow 4}}\left(\frac{\log _5\left(\frac{5 x}{4}\right)}{\log _5 \frac{20}{x}}\right)^{\left(\log _5 \frac{x}{4}\right)^{-1}}

,then [L] is (where [.] denotes greatest integer function)

View answer and explanation

Correct option: B

Option B (7) is correct because the limit L evaluates to

e^2

, and the greatest integer function of

e^2

is 7.

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16 Easy

16 of 930

Analyzing +4 | -1

\lim \limits_ {x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{\sqrt{\sin x}}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because by using the limit

\lim_{x \to 0} \frac{\sin x}{x} = 1

, the expression simplifies to 1.

17 Easy

17 of 930

Application +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} - 3,\;\;2 < x < 3}\\ {2x + 5,\;\;\;3 < x < 4} \end{array}} \right.

, the equation whose roots are

\lim\limits_{x \to {3^ - }} f\left( x \right)

and

\lim\limits_{x \to {3^ + }} f\left( x \right)

, is

View answer and explanation

Correct option: C

{x^2} - 17x + 66 = 0

Option C (

{x^2} - 17x + 66 = 0

) is correct because the roots of the equation are 6 and 11, which are the left-hand and right-hand limits, respectively, and the equation is formed correctly.

18 Easy

18 of 930

Application +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {7x - 3}&{{\text{if}}}&{x \geqslant 2} \\ {3x + 5}&{{\text{if}}}&{x < 2} \end{array}} \right.

then

\lim\limits_{x \to 2} f\left( x \right)

equals

View answer and explanation

Correct option: A

Option A (11) is correct because both the left-hand limit and the right-hand limit as x approaches 2 are equal to 11.

19 Medium

19 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{1 + \sin x - \cos x + \log \left( {1 - x} \right)}}{{{x^3}}}

View answer and explanation

Correct option: C

- \frac{1}{2}

Option C (-

\frac{1}{2}

) is correct because using Taylor series expansions for the trigonometric and logarithmic functions, the limit evaluates to -

\frac{1}{2}

20 Easy

20 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}

equals.

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because after applying trigonometric identities and simplifying, the limit evaluates to

\frac{1}{2}

21 Medium

21 of 930

Application +4 | -1

{l_1} = \lim\limits_{x \to - 2} \left( {x + \left| x \right|} \right),\;{l_2} = \lim\limits_{x \to - 2} \left( {2x + \left| x \right|} \right)

and

{l_3} = \lim\limits_{x \to \pi /2} \frac{{\cos x}}{{x - \pi /2}}

, then

View answer and explanation

Correct option: B

{l_2} < {l_3} < {l_1}

Option B (

{l_2} < {l_3} < {l_1}

) is correct because it accurately compares the limits calculated by considering the absolute value and the limit of the trigonometric function.

22 Easy

22 of 930

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x}\;\;\;\;\;x \ne 0}\\ {0\;\;\;\;\;\;x = 0} \end{array}} \right.

, then

\lim\limits_{x \to 0} f\left( x \right)

is equal to

View answer and explanation

Correct option: B

Option A (

\left( { - 6\;, - 3} \right) \cup \left( {3,6} \right)

) is correct because it represents the values of

x

that satisfy the inequality

3 < |x| < 6

23 Medium

23 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left[\frac{\sin (\operatorname{sgn}(x))}{\operatorname{sgn}(x)}\right]

, where [·] denotes the greatest integer function is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the limit from both the right and left sides equals 0, considering the greatest integer function.

24 Easy

24 of 930

Analyzing +4 | -1

\lim\limits_{x \to 1} \left( {\frac{{\int\limits_0^{{{(x - 1)}^2}} {t\cos ({t^2})dt} }}{{(x - 1)\sin (x - 1)}}} \right)

View answer and explanation

Correct option: D

Option D is correct because by applying L'Hopital's rule twice, the limit evaluates to 0.

25 Easy

25 of 930

Analyzing +4 | -1

Let

\left[ t \right]

denote the greatest integer

\le t

. If for some

\lambda \in R - \left\{ {0,\,1} \right\}

\lim\limits_{x \to 0} \left| {\frac{{1 - x + |x|}}{{\lambda - x + \left[ x \right]}}} \right| = L

, then

L

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because by evaluating the limit from both sides and equating them, we find L = 2.

26 Medium

26 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x}-\cot x\right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because after applying L'Hôpital's rule twice, the limit evaluates to 0.

27 Easy

27 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 1}} \frac{x+x^2+\ldots+x^n-n}{x-1}

, is

View answer and explanation

Correct option: C

\frac{n(n+1)}{2}

Option D (

\frac{\sqrt{3}+\sqrt{2}}{4}

) is correct because dividing by x and evaluating the limit at infinity gives the correct result. The provided answer is the rationalized form of the correct answer.

28 Medium

28 of 930

Application +4 | -1

\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2

, then the values of

a

and

b

, are

View answer and explanation

Correct option: B

a=1, b \in \mathrm{R}

Option B (a=1, b ∈ R) is correct because the limit depends only on 'a', and 'b' can take any real value.

29 Medium

29 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because after rationalizing and taking the limit as x approaches infinity, the expression simplifies to 1/2.

30 Easy

30 of 930

Analyzing +4 | -1

\lim\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}}

View answer and explanation

Correct option: A

exists and equals

\frac{1}{{4\sqrt 2 }}

Option A (exists and equals 1/(4√2)) is correct because after rationalizing the numerator twice and simplifying, the limit can be evaluated directly.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 930 single-correct-answer MCQs on LIMITS.
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