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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 361-390 of 930 questions

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361 Medium

361 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{\log x^n-[x]}{[x]}

, where

n \in N

and [.] denotes the greatest integer function, is

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the limit of the given expression as x approaches infinity simplifies to -1.

362 Medium

362 of 930

Application +4 | -1

\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=

View answer and explanation

Correct option: B

e^5

Option B (

e^5

) is correct because the limit can be evaluated using the standard limit formula

\lim_{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a

363 Easy

363 of 930

Understanding +4 | -1

The value of

\lim \limits_{{x \rightarrow-\infty}}\left[\tan ^{-1} x\right]

View answer and explanation

Correct option: A

-2

Option A (-2) is correct because as x approaches negative infinity,

\tan^{-1}x

approaches -

\frac{{\pi}}{{2}}

, and the greatest integer function of a value approaching -

\frac{{\pi}}{{2}}

is -2, since -

\frac{{\pi}}{{2}} \approx -1.57

364 Hard

364 of 930

Analyzing +4 | -1

g\left( x \right)

is a polynomial satisfying

g\left( x \right)g\left( y \right) = g\left( x \right) + g\left( y \right) + g\left( {xy} \right) - 2

for all real

x

and

y

and

g\left( 2 \right) = 5

, then

\lim\limits_{x \to 3} g\left( x \right)

View answer and explanation

Correct option: B

Option B (10) is correct because by substituting the given condition into the functional equation and solving for g(x), we find that g(x) = x² + 1, and thus

\lim\limits_{x \to 3} g(x) = 10

365 Medium

365 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to a} \frac{{\log \left( {x - a} \right)}}{{\log \left( {{e^x} - {e^a}} \right)}}

, is

View answer and explanation

Correct option: A

Option A (1) is correct because applying L'Hopital's rule or using the definition of the derivative, the limit evaluates to 1.

366 Easy

366 of 930

Application +4 | -1

\lim _{x \rightarrow+\infty}\left[2^{x-1} \tan \left(\frac{3}{2^x}\right)\right]=

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because, after using the limit property

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

, the limit of the expression as x approaches infinity is 3/2.

367 Medium

367 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e^5

Option B (

e^5

) is correct because by using the limit property

\lim_{x \to \infty} (1 + \frac{{a}}{{x}})^x = e^a

, the limit evaluates to

e^5

368 Medium

368 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow a}}\left\{\left[\left(\frac{a^{1 / 2}+x^{1 / 2}}{a^{1 / 4}-x^{1 / 4}}\right)^{-1}\right.\right.\left.-\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}\left.-\sqrt{2}^{\log_4a}\right\}^8

is equal to

View answer and explanation

Correct option: C

a^2

Option C (a²) is correct because after simplifying the expression and substituting x=a, the limit evaluates to a².

369 Easy

369 of 930

Analyzing +4 | -1

The value of

\lim\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}

, where

x < - 1

View answer and explanation

Correct option: C

Option C (1) is correct because, by dividing both the numerator and denominator by x^n, the limit simplifies to 1 as n approaches infinity, given that x < -1.

370 Easy

370 of 930

Application +4 | -1

The magnitude of the two forces forming a couple is

36 {~N}

each and the arm of the couple is

4 {~m}

. The magnitude of each force of an equivalent couple whose arm

9 {~m}

, is (in Newtons)

View answer and explanation

Correct option: C

Option C (16) is correct because the moment of a couple remains constant, so the force must adjust inversely with the arm length.

371 Easy

371 of 930

Application +4 | -1

\lim\limits_{x \to \frac{\pi }{6}} \,\,\frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}

is equal to

View answer and explanation

Correct option: B

- 3

Option B (-3) is correct because the limit is evaluated using L'Hopital's rule or by factoring and simplifying the expression.

372 Medium

372 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{3}

Option A (

\frac{\pi}{3}

) is correct because the limit can be evaluated by rationalizing the expression and then applying the inverse cosine function.

373 Easy

373 of 930

Analyzing +4 | -1

\lim _{x \rightarrow-\infty}\left(\sqrt{2 x^2-3}-5 x\right)

View answer and explanation

Correct option: B

+\infty

Option B (+∞) is correct because as x approaches negative infinity, the expression tends towards positive infinity.

374 Easy

374 of 930

Application +4 | -1

f'\left( 2 \right) = 2,f''\left( 2 \right) = 1

, then

\lim\limits_{x \to 2} \frac{{2{x^2} - 4f'\left( x \right)}}{{x - 2}}

, is

View answer and explanation

Correct option: A

Option A (4) is correct because applying L'Hopital's rule and using the given values of f'(2) and f''(2) results in this value.

375 Easy

375 of 930

Analyzing +4 | -1

0 < a < b

, then

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

View answer and explanation

Correct option: B

Equals

- 1

Option B (-1) is correct because as n approaches infinity, (a/b)^n approaches 0, simplifying the expression.

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376 Easy

376 of 930

Analyzing +4 | -1

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

equals

View answer and explanation

Correct option: B

\pi

Option B (

\pi

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

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

377 Easy

377 of 930

Application +4 | -1

Let

a=\min \left\{x^2+2 x+3: x \in R\right\}

and

b=\lim \limits_{{\theta \rightarrow 0}} \frac{1-\cos \theta}{\theta^2}

. Then, the value of

\sum\limits_{{r=0}} a^r b^{n-r}

, is

View answer and explanation

Correct option: C

\frac{4^{n+1}-1}{3 \times 2^n}

Option C,

\frac{4^{n+1}-1}{3 \times 2^n}

, is correct because it is the result of simplifying the expression and using the formula for the sum of a geometric series.

378 Medium

378 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}} \frac{1+2-3+4+5-6+\ldots+(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^{4}+4 n+3}-\sqrt{n^{4}+5 n+4}}

is :

View answer and explanation

Correct option: B

\frac{3}{2}(\sqrt{2}+1)

Option B (

\frac{3}{2}(\sqrt{2}+1)

) is correct because we simplify the numerator and denominator separately, then rationalize the denominator and take the limit as n approaches infinity.

379 Easy

379 of 930

Application +4 | -1

Let

L=\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^3} \sum\limits_{{k=1}}^n k^2 e^{\frac{k}{n}}

, then find the value of

e-L

View answer and explanation

Correct option: A

Option A (2) is correct because the limit L is evaluated as a definite integral, and then e-L is calculated.

380 Easy

380 of 930

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: A

{e^{ - 2/3}}

Option A (

{e^{ - 2/3}}

) is correct because after simplifying the expression and applying the limit, the result is

{e^{ - 2/3}}

381 Easy

381 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because the limit is evaluated using trigonometric identities and standard limit results, resulting in 2.

382 Easy

382 of 930

Analyzing +4 | -1

If for some real number

k

\lim\limits_{x \to 0} \,\,kx\,\,\mathrm{cosec}\left( x \right) = \lim\limits_{x \to 0} \,\,x\mathrm{cosec}\left( {kx} \right)

, then the possible values of

k

are

View answer and explanation

Correct option: A

1, - 1

Option A (

1, -1

) is correct because these are the values of k that satisfy the given limit equation after simplification.

383 Easy

383 of 930

Application +4 | -1

The value of

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

is equal to

View answer and explanation

Correct option: C

{e^{2/\pi }}

Option C (

{e^{2/\pi }}

) is correct because the limit can be evaluated by using L'Hopital's rule after rewriting the expression.

384 Easy

384 of 930

Analyzing +4 | -1

\alpha > \beta > 0

are the roots of the equation

a x^{2}+b x+1=0

, and

\lim \limits_{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)

, then

k

is equal to

View answer and explanation

Correct option: C

2 \alpha

Option C (2α) is correct because by using L'Hopital's rule and the relationship between the roots and coefficients of the quadratic equation, we can find the value of k.

385 Medium

385 of 930

Analyzing +4 | -1

\lim\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{{{\left( {x - 1} \right)}^2}}} = 2,\;

then (a, b, c) is

View answer and explanation

Correct option: A

\left( {2, - 4,2} \right)

Option A ((2, -4, 2)) is correct because for the limit to exist, the numerator must be equivalent to 2(x-1)², leading to a=2, b=-4, and c=2.

386 Easy

386 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(1+x \log \left(1+b^2\right)\right)^{1 / 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 equating it to 2bsin²θ, we find that sin²θ = 1, which implies θ =

\pm \frac{\pi}{2}

387 Easy

387 of 930

Application +4 | -1

Evaluate

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

View answer and explanation

Correct option: C

1/2

Option C (1/2) is correct because after simplifying the expression and applying the limit, the result is 1/2.

388 Medium

388 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}(\sin x)^{\tan x}

View answer and explanation

Correct option: A

Option A (1) is correct because after applying L'Hôpital's rule to the logarithm of the function, the limit evaluates to 0, and thus the original limit is e⁰ = 1.

389 Easy

389 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{x\left(e^{x}-1\right)}{1-\cos x}=

View answer and explanation

Correct option: D

Option D (2) is correct because applying L'Hopital's rule or using series expansions, the limit evaluates to 2.

390 Easy

390 of 930

Application +4 | -1

f\left( 9 \right) = 9,\;f'\left( 9 \right) = 4

, then

\lim\limits_{x \to 9} \frac{{\sqrt {f\left( x \right)} - 3}}{{\sqrt x - 3}}

equals

View answer and explanation

Correct option: A

The correct answer is A (4) because applying L'Hopital's rule and using the given values of f(9) and f'(9) leads to the solution.

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