Given that $\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}$, is

${\left( {\frac{e}{4}} \right)^k}$

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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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451 Easy

451 of 930

Application +4 | -1

\lim\limits_{x \to 0} \left[ {\frac{{{2^x} - 1}}{{\sqrt {1 + x} - 1}}} \right]

is equal to

View answer and explanation

Correct option: C

{\log _e}4

Option C (

\log_e 4

) is correct because by applying L'Hopital's rule, the limit evaluates to ln(4).

452 Easy

452 of 930

Analyzing +4 | -1

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

, when

\alpha \in Q

is equal to

View answer and explanation

Correct option: C

e^{1-\alpha}

Option C (

e^{1-\alpha}

) is correct because the limit can be evaluated by using the binomial expansion and the limit definition of e.

453 Medium

453 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left[1+x \log \left(1+b^2\right)\right]^{(1 i x)}=2 b \sin ^2 \theta, b > 0

and

\theta \in(-\pi, \pi]

then the value of

\theta

View answer and explanation

Correct option: D

\pm \frac{\pi}{2}

Option D (

\pm \frac{\pi}{2}

) is correct because by evaluating the limit and solving the resulting equation, the value of

\theta

is found to be

\pm \frac{\pi}{2}

454 Easy

454 of 930

Application +4 | -1

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

is equal to:

View answer and explanation

Correct option: C

\frac{3}{2}

Option C (

\frac{3}{2}

) is correct because it is the value of the limit after applying L'Hopital's Rule twice.

455 Easy

455 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{\sin {x^n}}}{{{{\left( {\sin x} \right)}^m}}},\;(m < n)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because, as x approaches 0, the expression simplifies to x^(n-m), and since m < n, the limit is 0.

456 Easy

456 of 930

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 is evaluated by applying L'Hopital's rule or by using trigonometric identities and standard limit formulas.

457 Easy

457 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(k-2) x-2 k\right\}}{x^2-4 x+4}=5

then

k

is equal to:

View answer and explanation

Correct option: D

Option D (3) is correct because the limit can be evaluated by factoring and using the standard limit for tan(x)/x.

458 Hard

458 of 930

Application +4 | -1

Given that

\lim\limits_{n \to \infty } \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\lim\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}

, is

View answer and explanation

Correct option: C

{\left( {\frac{4}{e}} \right)^k}

Option C (

{\left( {\frac{4}{e}} \right)^k}

) is correct because the limit can be expressed as an integral and evaluated using the given information.

459 Easy

459 of 930

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\sin 4x}}{{1 - \sqrt {1 - x} }}

, is

View answer and explanation

Correct option: B

Option B (8) is correct because the limit can be evaluated by rationalizing the denominator and using the standard limit

\lim_{x \to 0} \frac{{\sin ax}}{x} = a

460 Easy

460 of 930

Understanding +4 | -1

Let

f(x)=\langle x\rangle^*

, where

\langle x\rangle^*

is the distance from

x

to the integer nearest to

x

then

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

View answer and explanation

Correct option: C

Option C (0) is correct because as x approaches 2, the distance from x to the nearest integer approaches 0.

461 Medium

461 of 930

Understanding +4 | -1

Which of the following is/are true?

I.

\lim \limits_{{x \rightarrow 0}} \frac{\sin x}{x}=1

II.

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

View answer and explanation

Correct option: A

Both I and II are true

Option A (Both I and II are true) is correct because both given limits are standard results in calculus.

462 Medium

462 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)=

View answer and explanation

Correct option: B

\frac{1}{2}

Option C (1/4) is correct because simplifying the expression using the sum of cubes formula leads to a limit of 1/4 as n approaches infinity.

463 Easy

463 of 930

Application +4 | -1

\lim\limits_{x \to 0} \left[ {\frac{{\sqrt {a + x} - \sqrt {a - x} }}{{4x}}} \right]

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by rationalizing the numerator and simplifying, the limit can be evaluated to

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

464 Easy

464 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{\log x}{\tan \pi x}

is equal to

View answer and explanation

Correct option: B

\frac{1}{\pi}

The correct answer is B (

\frac{1}{\pi}

) because applying L'Hôpital's rule to the limit yields the value

\frac{1}{\pi}

465 Medium

465 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} {\left\{ {\frac{{1 + \tan x}}{{1 + \sin x}}} \right\}^{\mathrm{cosec}x}}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because by rewriting the expression and using the limit definition of e, the limit simplifies to 1.

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466 Medium

466 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{4}{{{x^8}}}\left( {1 - \cos \frac{{{x^2}}}{2} - \cos \frac{{{x^2}}}{4} + \cos \frac{{{x^2}}}{2} \cdot \cos \frac{{{x^2}}}{4}} \right) = {\left( {\frac{1}{{{n_1}}}} \right)^n}

Where

{n_1}

is a Prime number, then value of

n_1+n

equals

View answer and explanation

Correct option: A

The correct answer is A. By simplifying the limit using trigonometric identities and standard limit results, the value of

n_1 + n

is found to be 8.

467 Easy

467 of 930

Analyzing +4 | -1

\alpha

and

\beta

are the roots of the equation

a x^2+b x+c=0

, then

\lim \limits_{{x \rightarrow \alpha}}\left(1+a x^2+b x+c\right)^{1 / x-\alpha}

, is

View answer and explanation

Correct option: A

e^{a(\alpha-\beta)}

Option B (

e^{a(\beta-\alpha)}

) is correct because the limit can be evaluated by factoring the quadratic and using exponential limit properties.

468 Easy

468 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \pi / 2}}\left[\sin ^{-1} \sin x\right]

View answer and explanation

Correct option: A

Option A (1) is correct because the greatest integer function of

\sin^{-1}(\sin x)

as x approaches

\pi/2

is 1.

469 Easy

469 of 930

Understanding +4 | -1

The value of

\lim\limits_{x \to \infty } \frac{{\sin x}}{x}

, is

View answer and explanation

Correct option: B

Option B (0) is correct because as x approaches infinity, sin(x) oscillates between -1 and 1, and dividing by an infinitely large x results in a limit of 0.

470 Easy

470 of 930

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: D

\frac{3}{2}

Option D (

\frac{3}{2}

) is correct because by applying L'Hopital's rule twice, the limit evaluates to 3/2.

471 Easy

471 of 930

Analyzing +4 | -1

Let

f: R \rightarrow[0, \infty)

be such that

\lim \limits_{{x \rightarrow 5}} f(x)

exists and

\lim \limits_{{x \rightarrow 5}} \frac{(f(x))^2-9}{\sqrt{|x-5|}}=0

. Then,

\lim \limits_{{x \rightarrow 5}} f(x)

equals

View answer and explanation

Correct option: C

Option C (3) is correct because the limit of f(x) as x approaches 5 must be 3, given the provided condition and the non-negativity of f(x).

472 Easy

472 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \log _{\sin x}(\sin 2 x)

is equal to

View answer and explanation

Correct option: B

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

473 Easy

473 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{3 x^2+2}{7 x^2+2}\right)^{1 / x^2}

is equal to

View answer and explanation

Correct option: A

e^{-2}

The correct answer is A (

e^{-2}

) because after applying the limit and simplifying, the expression converges to

e^{-2}

474 Easy

474 of 930

Analyzing +4 | -1

f(x)=\frac{(2 x+3)(\sqrt{x}-1)}{2 x^2+x-3}

then

\lim _{x \rightarrow 1} f(x)

is equal to

View answer and explanation

Correct option: D

-\frac{1}{10}

Option D (-1/10) is correct because after simplifying the expression and taking the limit as x approaches 1, the result is -1/10.

475 Easy

475 of 930

Application +4 | -1

\lim\limits_{x \to 0} \frac{d}{{dx}}\int \frac{{1 - \cos x}}{{{x^2}}}dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because the derivative and integral cancel out, leaving the limit of the original function, which evaluates to 1/2.

476 Easy

476 of 930

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\mathrm{when}\,\,x\,\,\mathrm{is }\,\mathrm{rational}}\\ {0,\mathrm{when}\,\,x\,\,\,\mathrm{irrational}} \end{array}} \right.

, then

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

View answer and explanation

Correct option: B

Option D (None of these) is correct because the limit does not exist as the left-hand limit and right-hand limit are not equal when approaching 0.

477 Easy

477 of 930

Understanding +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{|x|}{x}=

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left-hand limit and the right-hand limit are not equal.

478 Medium

478 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left(\sin ^4 x+\frac{1}{4} \sin ^4 2 x+\ldots \cdot \frac{1}{4^n} \sin ^4\left(2^n x\right)\right)

View answer and explanation

Correct option: A

\sin ^2 x

The correct answer is

\sin^2 x

because the limit of the given series telescopes to

\sin^2 x

as n approaches infinity.

479 Medium

479 of 930

Analyzing +4 | -1

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

equals

View answer and explanation

Correct option: A

\frac{1-\log a}{1+\log a}

Option A ((1-log a)/(1+log a)) is correct because after applying L'Hôpital's rule, the limit simplifies to (1-log a)/(1+log a).

480 Easy

480 of 930

Application +4 | -1

\alpha=\lim \limits_{{n \rightarrow \infty}}\left(\frac{1^2+2^2+\cdots+n^2}{n^3}\right)

and

\beta=\lim \limits_{{n \rightarrow \infty}}\left[\frac{\left(1^3-1^2\right)+\left(2^3-2^2\right)+\left(3^3-3^2\right)+\cdots+\left(n^3-n^2\right)}{n^4}\right]

Then

View answer and explanation

Correct option: D

3 \alpha=4 \beta

Option D (

3 \alpha = 4 \beta

) is correct because after evaluating the limits, the relationship between

\alpha

and

\beta

satisfies this equation.

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