$\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left\{ {\mathop \smallint \limits_y^a {e^{{{\sin }^2}t}}dt - \mathop \smallint \limits_{x + y}^a {e^{{{\sin }^2}t}}dt} \right\}$ is equal to (where $a$ is a constant)

$\sin \,2y\,\,{e^{{{\sin }^2}y}}$

$\mathop {\lim }\limits_{h \to 0} \frac{{\sin \sqrt {x + h} - \sin \sqrt x }}{h}$ is equal to

$1/\left( {2\sin \sqrt x } \right)$

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}$, is

$\log \left( {\frac{b}{a}} \right)$

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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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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 451-480 of 647 questions

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451 Easy

451 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\tan x-\sin x}{x^3}

, is

View answer and explanation

Correct option: A

\frac{1}{2}

The correct answer is A (1/2) because by rewriting the expression and using trigonometric identities and limits, the value is found to be 1/2.

452 Easy

452 of 647

Application +4 | -1

\lim\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^8} - 1}}{{{{\left( {1 + x} \right)}^2} - 1}}

is equal to

View answer and explanation

Correct option: C

Option C (4) is correct because after factoring and simplifying the expression, the limit as x approaches 0 is 4.

453 Easy

453 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow-2}} \frac{\left(x^2-x-6\right)^2}{(x+2)^2}

, is

View answer and explanation

Correct option: B

Option B (25) is correct because it is the result of factoring and simplifying the limit expression as x approaches -2.

454 Easy

454 of 647

Application +4 | -1

\operatorname{Lim}_{x \rightarrow 0} \frac{\ln (3+x)-\ln (3-x)}{x}=k

, the value of

k

View answer and explanation

Correct option: A

\frac{2}{3}

Option A (

\frac{2}{3}

) is correct because it is the limit of the given expression as x approaches 0, evaluated using L'Hôpital's rule or series expansion.

455 Easy

455 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{1}{x}\left\{ {\int \limits_y^a {e^{{{\sin }^2}t}}dt - \int \limits_{x + y}^a {e^{{{\sin }^2}t}}dt} \right\}

is equal to (where

a

is a constant)

View answer and explanation

Correct option: A

{e^{{{\sin }^2}y}}

Option A (

{e^{{{\sin }^2}y}}

) is correct because it is the result of applying the Leibniz rule for differentiating under the integral sign and then evaluating the limit.

456 Easy

456 of 647

Application +4 | -1

\lim _{x \rightarrow+\infty} \frac{\sqrt{2 x^2+3}}{4 x+3}=

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because after dividing by x and simplifying, the limit of the expression as x approaches infinity is

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

457 Easy

457 of 647

Application +4 | -1

\lim _{x \rightarrow 3} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}}

is equal to:

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the value of the limit after rationalizing and simplifying the expression.

458 Easy

458 of 647

Analyzing +4 | -1

Let

f(x)

be a polynomial function such that

f(x) + f'(x) + f''(x) = {x^5} + 64.

Then, the value of

\lim\limits_{x \to 1} \frac{{f(x)}}{{x - 1}}

is equal to

View answer and explanation

Correct option: A

– 15

Option B (– 60) is correct because by substituting a general 5th-degree polynomial and using L'Hopital's rule, the limit can be found to be -15.

459 Easy

459 of 647

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {a^x}\sin \left( {\frac{b}{{{a^x}}}} \right)

(a > 1)

View answer and explanation

Correct option: C

b

Option C (

b

) is correct because as

x

approaches infinity,

\frac{b}{{{a^x}}}

approaches 0, and the limit can be evaluated using the small angle approximation for sine.

460 Easy

460 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow a}} \frac{x^m-a^m}{x^n-a^n}

, is

View answer and explanation

Correct option: A

\frac{m}{n} a^{m-n}

Option A (

\frac{m}{n} a^{m-n}

) is correct because applying L'Hopital's rule or using the standard limit formula yields the given result.

461 Easy

461 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \ldots \sqrt[n]{\cos n x}}{x^2}

has the value equal to 10 , the value of

n

View answer and explanation

Correct option: A

The correct answer is A (6) because by using L'Hopital's rule and simplifying the expression, the value of n can be determined by solving the resulting quadratic equation.

462 Easy

462 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 1}}\left(\log _3 3 x\right)^{\log _x 3}

, is

View answer and explanation

Correct option: A

e

Option A (

e

) is correct because the limit can be simplified using logarithm properties and then evaluated using the exponential form of limits.

463 Easy

463 of 647

Application +4 | -1

\lim\limits_{x \to - \infty } \left( {3x + \sqrt {9{x^2} - x} } \right)

equals

View answer and explanation

Correct option: B

\frac{1}{6}

Option B (

\frac{1}{6}

) is correct because after rationalizing and simplifying, the limit evaluates to 1/6 as x approaches negative infinity.

464 Easy

464 of 647

Understanding +4 | -1

Let

f(x)= \begin{cases}\frac{\sin \left(x^2\right)}{x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{cases}

Then

\lim _{x \rightarrow 0} f(x)

View answer and explanation

Correct option: B

Option B is correct because as x approaches 0, x² also approaches 0, and sin(x²)/x² approaches 1. Therefore, the limit becomes 0 * 1 = 0.

465 Easy

465 of 647

Application +4 | -1

f(x)=\left\{\begin{array}{ll}3 x-1, & x \geq 1 \\ 2 x+3, & x < 1\end{array}, g(x)=\left\{\begin{array}{cc}3-x, & x < 2 \\ 2 x-3, & x \geq 2\end{array}\right.\right.

, then

\lim \limits_{{x \rightarrow 2}} f(g(x))=

View answer and explanation

Correct option: B

Option B (2) is correct because the limit of

f(g(x))

x

approaches 2 is 2, based on the piecewise definitions of

f(x)

and

g(x)

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466 Easy

466 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{x^n \sin ^n x}{x^n-\sin ^n x}

is non-zero finite, then

n

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because after applying L'Hopital's rule multiple times, the limit becomes finite and non-zero only when n=2.

467 Easy

467 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to 2} \frac{{x\left( {{5^x} - 1} \right)}}{{1 - \cos x}}

, is

View answer and explanation

Correct option: B

2\log 5

Option B (

2\log 5

) is correct because by applying L'Hopital's rule and limit properties, the expression simplifies to

2\log 5

468 Easy

468 of 647

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1+2+3+\ldots+n}{1+3+5+\ldots+(2 n-1)}=

View answer and explanation

Correct option: C

1 / 2

Option A (1/5) is correct because evaluating both limits separately and subtracting them gives 1/5.

469 Easy

469 of 647

Application +4 | -1

\lim\limits_{h \to 0} \frac{{\sin \sqrt {x + h} - \sin \sqrt x }}{h}

is equal to

View answer and explanation

Correct option: C

\left( {\cos \sqrt x } \right)/2\sqrt x

Option C (

\left( {\cos \sqrt x } \right)/2\sqrt x

) is correct because applying L'Hopital's rule to the limit results in this expression.

470 Easy

470 of 647

Application +4 | -1

\lim\limits_{x \to \infty } \frac{{{{\left( {2x + 1} \right)}^{40}}{{\left( {4x - 1} \right)}^5}}}{{{{\left( {2x + 3} \right)}^{45}}}}

is equal to

View answer and explanation

Correct option: C

Option C (32) is correct because after factoring out

x

and simplifying, the limit evaluates to 32.

471 Easy

471 of 647

Analyzing +4 | -1

\lim \limits_{{h \rightarrow 0}}\left\{\frac{1}{h \cdot \sqrt[3]{8+h}}-\frac{1}{2 h}\right\}

is equal to

View answer and explanation

Correct option: D

-\frac{1}{48}

Option D (-1/48) is correct because after simplifying the expression and using L'Hopital's rule or series expansion, the limit evaluates to -1/48.

472 Easy

472 of 647

Analyzing +4 | -1

The value of

\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}

View answer and explanation

Correct option: C

5 \sqrt{2}

The correct answer is C (

5 \sqrt{2}

) because by using L'Hopital's rule and trigonometric identities, the limit can be evaluated to

5\sqrt{2}

473 Easy

473 of 647

Application +4 | -1

\lim _{x \rightarrow 0} \frac{\sin 2 x}{x \cos x}

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because simplifying the expression using the double angle formula for sine and the limit of sin(x)/x as x approaches 0 leads to the answer 2.

474 Easy

474 of 647

Analyzing +4 | -1

The quadratic equation whose roots are the minimum value of

\sin ^2 \theta-\sin \theta+\frac{1}{2}

and

\lim \limits_{{x \rightarrow \infty}} \sqrt{(x+1)(x+2)}-x

View answer and explanation

Correct option: B

8 x^2-14 x+3=0

Option B,

8x^2 - 14x + 3 = 0

, is correct because the minimum value of the given trigonometric expression is

\frac{1}{4}

, and the limit of the given expression is

\frac{3}{2}

. Therefore, the quadratic equation with roots

\frac{1}{4}

and

\frac{3}{2}

8x^2 - 14x + 3 = 0

475 Easy

475 of 647

Application +4 | -1

\lim\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}

is equal to

View answer and explanation

Correct option: A

\frac{3}{2}

Option A (

\frac{3}{2}

) is correct because the limit can be evaluated by using L'Hopital's rule or by using Taylor series expansions for

e^{x^2}

and

\cos x

476 Easy

476 of 647

Application +4 | -1

The value of

\lim\limits_{x \to b} \frac{{\sqrt {x - a} - \sqrt {b - a} }}{{{x^2} - {b^2}}},

for

b>a

, is

View answer and explanation

Correct option: B

\frac{1}{4 b \sqrt{b-a}}

Option B (

\frac{1}{4 b \sqrt{b-a}}

) is correct because rationalizing the numerator and simplifying the expression leads to the given result.

477 Easy

477 of 647

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}

, is

View answer and explanation

Correct option: A

\log \left( {\frac{a}{b}} \right)

Option A (

\log \left( {\frac{a}{b}} \right)

) is correct because it's a direct application of the limit definition of the derivative of an exponential function.

478 Easy

478 of 647

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} + 6}}{{{x^2} - 6}}} \right)^x}

is given by

View answer and explanation

Correct option: B

Option B (1) is correct because the limit can be evaluated by rewriting the expression and using the standard limit definition of

e

479 Easy

479 of 647

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{{\sin }^2}2x}}{{{{\sin }^2}4x}}

View answer and explanation

Correct option: C

\frac{1}{4}

Option C (1/4) is correct because by using the property

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

, the limit evaluates to 1/4.

480 Easy

480 of 647

Application +4 | -1

\lim _{x \rightarrow 0}\left(\frac{\log (6+x)-\log (6-x)}{x}\right)=

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because, after applying logarithmic properties and the limit

\lim_{x \to 0} \frac{\log(1+x)}{x} = 1

, the limit of the expression as x approaches 0 is 1/3.

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