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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 481-510 of 647 questions

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481 Easy

481 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}

is equal to

View answer and explanation

Correct option: B

\pi

Option B (

\pi

) is correct because the limit can be evaluated by using the trigonometric identity

\sin(\pi - x) = \sin x

and the standard limit

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

482 Easy

482 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4

, then

View answer and explanation

Correct option: B

a=1, b=-4

Option B (

a=1, b=-4

) is correct because for the limit to be 4, the coefficients of

x^2

must be 0 and the ratio of the coefficients of

x

must be 4.

483 Easy

483 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because the limit simplifies to

\pi

after applying trigonometric identities and limit properties.

484 Easy

484 of 647

Application +4 | -1

Let

f:R \to R

be a differentiable function such that

f\left( 3 \right) = 3,f'\left( 3 \right) = \frac{1}{2}

, Then, the value of

\lim\limits_{x \to 3} \,\frac{{\int \nolimits_3^{f\left( x \right)} 2{t^3}\,dt}}{{x - 3}}

View answer and explanation

Correct option: C

Option C (27) is correct because applying L'Hopital's rule and the given values of f(3) and f'(3) leads to the result 27.

485 Easy

485 of 647

Analyzing +4 | -1

The value

\lim\limits_{x \to a} {\left( {\frac{{\sin x}}{{\sin a}}} \right)^{\frac{1}{{x - a}}}}

, is

View answer and explanation

Correct option: C

{e^{\cot a}}

Option C (

{e^{\cot a}}

) is correct because it follows from applying L'Hopital's rule after rewriting the limit in exponential form.

486 Easy

486 of 647

Understanding +4 | -1

Let

f(x)=\operatorname{sgn}(\operatorname{sgn}(\operatorname{sgn} x))

. Then

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand limit and right-hand limit are not equal, so the limit does not exist.

487 Easy

487 of 647

Understanding +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{a^n+b^n}{a^n-b^n}

, where

a > b > 1

, is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because dividing both numerator and denominator by

a^n

and using the fact that

(\frac{{b}}{{a}})^n

approaches 0 as n approaches infinity (since a > b > 1), the limit simplifies to 1.

488 Easy

488 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \,\,\frac{{{{\left( {\int \nolimits_0^x \,\,{e^x}\;\,dx} \right)}^2}}}{{\int \nolimits_0^x \,\,{e^{2{x^3}}}\;\,dx}}

, is

View answer and explanation

Correct option: D

\mathrm{0}

Option D (0) is correct because applying L'Hopital's rule repeatedly, the limit approaches 0 as the exponential term in the denominator grows much faster than the numerator.

489 Easy

489 of 647

Application +4 | -1

\lim\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}

View answer and explanation

Correct option: B

e

Option B (

e

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

e

490 Easy

490 of 647

Understanding +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{a x^2+b x+c}{d x^2+e x+f}

, is

View answer and explanation

Correct option: A

\frac{a}{d}

Option A (

\frac{a}{d}

) is correct because when x approaches infinity, the highest power terms dominate the limit.

491 Easy

491 of 647

Application +4 | -1

y_n=\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}

then

\lim _{n \rightarrow \infty} y_n

is equal to

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because the sum of the squares of the first n natural numbers is

\frac{n(n+1)(2n+1)}{6}

, and dividing by

n^3

and taking the limit as

n

approaches infinity yields

\frac{1}{3}

492 Easy

492 of 647

Application +4 | -1

For

x \in R\,\,\lim\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}

is equal to

View answer and explanation

Correct option: C

{e^{ - 5}}

Option C (

{e^{ - 5}}

) is correct because by rewriting the expression and using the limit definition of

e

, the limit evaluates to

{e^{ - 5}}

493 Easy

493 of 647

Application +4 | -1

\lim\limits_{x \to 1} \left( {1 + \cos \pi \mathrm{ }x} \right){\cot ^2}\pi

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 1/2.

494 Easy

494 of 647

Analyzing +4 | -1

The value of

\lim _{x \rightarrow 0^{+}} x^m(\log x)^n, m

\mathrm{n} \in \mathrm{N}

View answer and explanation

Correct option: A

Option A is correct. By applying L'Hopital's rule repeatedly, the limit can be evaluated to 0.

495 Easy

495 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} k x \operatorname{cosec} x=\lim \limits_{{x \rightarrow 0}} x \operatorname{cosec} k x

, then

k=

View answer and explanation

Correct option: C

\pm 1

Option C (

\pm 1

) is correct because solving the equation

\lim \limits_{{x \rightarrow 0}} k x \operatorname{cosec} x=\lim \limits_{{x \rightarrow 0}} x \operatorname{cosec} k x

yields

k = \pm 1

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496 Easy

496 of 647

Analyzing +4 | -1

x > 0

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

View answer and explanation

Correct option: C

Option C (

\sin 1

) is correct. The limit is found by evaluating the left-hand limit and right-hand limit separately and showing that they are equal to

\sin 1

497 Easy

497 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^n \cot ^{-1}\left(\frac{r^3-r+\frac{1}{r}}{2}\right)

View answer and explanation

Correct option: C

\frac{\pi}{2}

Option C (

\frac{\pi}{2}

) is correct because the given sum telescopes after applying the arctangent subtraction formula, and the limit as n approaches infinity is

\frac{\pi}{2}

498 Easy

498 of 647

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {{e^{kx}} - 1} \right)\sin kx}}{{{x^2}}} = 4

, then

k

is equal to

View answer and explanation

Correct option: C

\pm \,2

Option C (

\pm \,2

) is correct because solving the limit equation yields two possible values for k, both positive and negative 2.

499 Easy

499 of 647

Application +4 | -1

The value of

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

is:

View answer and explanation

Correct option: B

e^a

Option B (

e^a

) is correct because the limit can be evaluated using the standard limit formula for exponential functions and L'Hopital's rule.

500 Easy

500 of 647

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{x + 6}}{{x + 1}}} \right)^{x + 4}}

, is

View answer and explanation

Correct option: D

{e^5}

Option D (

{e^5}

) is correct because it is the result of evaluating the limit of the given expression as x approaches infinity, using the standard limit form for e.

501 Easy

501 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}|x|^{[\cos x]}

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because as x approaches 0,

\cos x

approaches 1, and any number (except 0) raised to the power of 0 is 1.

502 Easy

502 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{1-\cos ^3 x}{x \sin x \cos x}=

View answer and explanation

Correct option: C

3 / 2

Option C (

3 / 2

) is correct because the limit is evaluated by using trigonometric identities and standard limit results.

503 Easy

503 of 647

Analyzing +4 | -1

Let

f(x)

be a real valued function defined for all

x \geq 1

, satisfying

f(1)=1

and

f^{\prime}(x)=\frac{1}{x^2+(f(x))}

; then

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

View answer and explanation

Correct option: C

exists and less than

1+\frac{\pi}{4}

Option C, exists and less than

1 + \frac{\pi}{4}

, is correct because

f'(x) > 0

implies

f(x)

is increasing, and by comparing

f'(t)

with

\frac{1}{1+t^2}

and integrating, we find that

\lim_{x \to \infty} f(x) < 1 + \frac{\pi}{4}

504 Easy

504 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{1 + 3x}}{{2 + 3x}}} \right)^{\frac{{1 - \sqrt x }}{{1 + x}}}}

, is

View answer and explanation

Correct option: D

Option D (1) is correct because, as x approaches infinity, the base approaches 1 and the exponent approaches 0, resulting in a limit of 1.

505 Easy

505 of 647

Understanding +4 | -1

The value of

\lim\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)

View answer and explanation

Correct option: C

Option C (1) is correct because as x approaches infinity, the highest power terms dominate, and the ratio of the coefficients of x² in the numerator and denominator is 1.

506 Easy

506 of 647

Analyzing +4 | -1

\lim \limits_{{h \rightarrow 0}} \frac{e^x+e^{-x}-2 \cos x}{2 x \sin x}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because after applying L'Hôpital's rule twice, the limit evaluates to 1.

507 Easy

507 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{2 x^{1 / 2}+3 x^{1 / 3}+4 x^{1 / 4}+\ldots n x^{1 / n}}{(2 x-3)^{1 / 2}+(2 x-3)^{1 / 3}+\ldots+(2 x-3)^{1 / n}}

View answer and explanation

Correct option: C

\sqrt{2}

Option C (

\sqrt{2}

) is correct because it is obtained by dividing each term by

x^{1/2}

and taking the limit as x approaches infinity.

508 Easy

508 of 647

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

\frac{\log _e(e / a)}{\log _e(a e)}

Option B (

\frac{\log _e(e / a)}{\log _e(a e)}

) is correct because applying L'Hopital's rule and simplifying leads to this expression.

509 Easy

509 of 647

Application +4 | -1

\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0

then

(a, b)

value is

View answer and explanation

Correct option: B

(1,-1)

Option B ((1,-1)) is correct because, for the limit to be 0, the coefficients of

x^2

and

x

in the numerator must be zero, leading to

a=1

and

b=-1

510 Easy

510 of 647

Application +4 | -1

The value of

\lim \limits_{x \rightarrow \frac{\pi}{4}}(1+[x])^{\frac{1}{\ln (\tan x)}}

is:
(where [.] denotes greatest integer function).

View answer and explanation

Correct option: B

Option B (1) is correct because as x approaches

\frac{\pi}{4}

, the greatest integer function of x is 1, and any constant raised to a finite power is the constant itself.

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