Let $f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{2n}} - 1}}{{{x^{2n}} + 1}}$, then

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\left| x \right| > 1}\\ { - 1,\;\left| x \right| < 1} \end{array}} \right.$

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1,\;\left| x \right| 1} \end{array}} \right.$

$f\left( x \right)$ is not defined for any value of $x$

$f\left( x \right) = 1$ for $\left| x \right| = 1$

$\mathop {\lim }\limits_{x \to \pi /4} \,\,\frac{{\mathop \smallint \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - {\pi ^2}/16}}$ equals

$\frac{2}{\pi }f\left( 2 \right)$

$\frac{2}{\pi }f\left( {\frac{1}{2}} \right)$

Back to LIMITS

Your Progress

0 / 930 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 301-330 of 930 questions

Preparing practice and saved progress…

301 Easy

301 of 930

Application +4 | -1

\lim \limits_{{h \rightarrow 0}} \frac{\sqrt{x+h}-\sqrt{x}}{h}=

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the limit represents the derivative of

\sqrt{x}

302 Medium

302 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

The correct answer is B (0) because as n approaches infinity, the terms (2^(1/2) - 2^(1/(2n+1))) approach (2^(1/2) - 1), and since each term is less than 1, the infinite product converges to 0.

303 Hard

303 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17

, then

5 a^{2}+b^{2}

is equal to

View answer and explanation

Correct option: C

Option C (68) is correct because it is obtained by using L'Hopital's rule and Taylor series expansion to evaluate the limit and then solving for the expression.

304 Easy

304 of 930

Application +4 | -1

\lim\limits_{x \to \infty } \frac{{\int \nolimits_0^{2x} x\;{e^{{x^2}}}\;dx}}{{{e^{4{x^2}}}}}

equals

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because after evaluating the integral and applying the limit, the result is

\frac{1}{2}

305 Easy

305 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{e^{3 x}-1}{\tan x}

View answer and explanation

Correct option: C

The correct answer is C (3) because applying L'Hôpital's rule to the limit yields the value 3.

306 Easy

306 of 930

Application +4 | -1

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

View answer and explanation

Correct option: A

e^4

Option A (

e^4

) is correct because after applying L'Hopital's rule and simplifying, the limit evaluates to

e^4

307 Easy

307 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \pi / 6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}=

View answer and explanation

Correct option: B

-3

Option B (-3) is correct because after factoring and simplifying the limit expression, substituting x = pi/6 yields -3.

308 Easy

308 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \,\frac{{1 - \cos \left( {1 - \cos x} \right)}}{{{x^4}}}

View answer and explanation

Correct option: D

\frac{1}{8}

Option D (1/8) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 1/8.

309 Easy

309 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to {0^ + }} \frac{{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})}}{{x - {x^3}}}

,Where [x] denotes the greatest integer

\le x

View answer and explanation

Correct option: D

\frac{\pi }{2}

Option D (

\frac{\pi}{2}

) is correct because, as x approaches 0 from the positive side, the expression simplifies to (arcsin(x)/x) * arccos(x), which evaluates to 1 * (

\frac{\pi}{2}

310 Easy

310 of 930

Analyzing +4 | -1

Let

f\left( x \right) = \lim\limits_{x \to \infty } \frac{{{x^{2n}} - 1}}{{{x^{2n}} + 1}}

, then

View answer and explanation

Correct option: A

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\;\left| x \right| > 1}\\ { - 1,\;\left| x \right| < 1} \end{array}} \right.

Option A (

f(x) = \begin{{cases}} 1, |x| > 1 \\ -1, |x| < 1 \end{{cases}}

) is correct because as n approaches infinity, x⁽²ⁿ⁾ approaches 0 when |x| < 1 and approaches infinity when |x| > 1, leading to the given piecewise function.

311 Medium

311 of 930

Application +4 | -1

\lim\limits_{x \to \pi /4} \,\,\frac{{\int \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - {\pi ^2}/16}}

equals

View answer and explanation

Correct option: A

\frac{8}{\pi }f\left( 2 \right)

Option A (

\frac{8}{\pi }f\left( 2 \right)

) is correct because applying Leibniz's rule and L'Hopital's rule results in this expression.

312 Medium

312 of 930

Analyzing +4 | -1

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

, is equal to

View answer and explanation

Correct option: A

1 / 3

Option A (1/3) is correct because the limit is evaluated by simplifying the summation expressions and taking the limit as n approaches infinity.

313 Easy

313 of 930

Application +4 | -1

\lim\limits_{x \to - \infty } \,\frac{{2x - 1}}{{\sqrt {{x^2} + 2x + 1} }}

is equal to

View answer and explanation

Correct option: B

- 2

Option B (-2) is correct because the limit as x approaches negative infinity simplifies to -2 after considering the sign of x in the square root.

314 Easy

314 of 930

Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{n^p \sin ^2(n !)}{n+1}, 0 < p < 1

, is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the expression can be simplified to an oscillating number divided by infinity, which approaches 0.

315 Easy

315 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^2+2 \sin x+1}-\sqrt{\sin ^2 x-x+1}}

View answer and explanation

Correct option: B

Option B (2) is correct because after rationalizing and dividing by x, the limit simplifies to 2 using the standard limit

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

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

316 Easy

316 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: D

\frac{2}{9}

Option D (

\frac{2}{9}

) is correct because after simplifying the expression and dividing by the highest power of x, the limit converges to

\frac{2}{9}

317 Easy

317 of 930

Application +4 | -1

\lim _{x \rightarrow 0} \frac{x+\sin 2 x}{x-\sin 2 x}

is equal to

View answer and explanation

Correct option: B

-3

The correct answer is B (-3) because applying L'Hôpital's rule to the limit yields the value -3.

318 Easy

318 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

e^{m k}

Option C (

e^{mk}

) is correct because the limit is a standard exponential limit of the form (1 + x/n)^(nx) which converges to e^x.

319 Easy

319 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{x^m-1}{x^n-1}

is equal to (

m

and

n

are positive integers)

View answer and explanation

Correct option: A

\frac{m}{n}

Option A (

\frac{m}{n}

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

\frac{m}{n}

320 Easy

320 of 930

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: A

{e^2}

Option A (

{e^2}

) is correct because the limit can be transformed into the form

{e^{\lim f(x)}}

and the exponent evaluates to 2.

321 Medium

321 of 930

Analyzing +4 | -1

f\left( x \right)

is differentiable function and

f''\left( 0 \right) = a

, then

\lim\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}

is equal to

View answer and explanation

Correct option: A

3a

Option A (

3a

) is the correct answer because it is obtained by applying L'Hopital's rule twice and then evaluating the limit.

322 Easy

322 of 930

Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{n \cdot 3^n}{n(x-2)^n+n \cdot 3^{n+1}-3^n}=\frac{1}{3}

, where

n \in N

, then the number of integer (s) in the range of

x

, is

View answer and explanation

Correct option: C

Option C (5) is correct because the limit condition leads to an inequality for x, resulting in 5 integer values within the range.

323 Easy

323 of 930

Application +4 | -1

\lim\limits_{x \to 1} \frac{{\sum \nolimits_{r = 1}^n \,\,{x^r} - n}}{{x - 1}}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the limit can be evaluated by recognizing it as the sum of the first n integers.

324 Medium

324 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{{e^{1/x}}}}{{{e^{1/x + 1}}}}

is equal to

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the limit depends on the direction of approach to 0, and the limit is 1/e.

325 Medium

325 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left\{1+x \ln \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

, is

View answer and explanation

Correct option: D

\pm \frac{\pi}{2}

Option D (

\pm \frac{\pi}{2}

) is correct because by simplifying the given limit equation, we find that

\sin^2 \theta = 1

, which leads to

\theta = \pm \frac{\pi}{2}

326 Medium

326 of 930

Analyzing +4 | -1

\lim\limits_{{\text{x}} \to 0} \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}

is equal to:

View answer and explanation

Correct option: C

\frac{1}{6}

Option C (

\frac{1}{6}

) is correct because the limit is evaluated using trigonometric identities and series expansion.

327 Easy

327 of 930

Application +4 | -1

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

View answer and explanation

Correct option: B

Option B (0) is correct because the limit of (sec x - tan x) as x approaches pi/2 is 0 after rewriting in terms of sine and cosine and simplifying.

328 Medium

328 of 930

Application +4 | -1

\lim\limits_{x \to 0} \,\frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}\,

is equal to

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because using trigonometric identities and limit properties, the given limit simplifies to

\pi

329 Medium

329 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{1+\sin x-\cos x+\log _e(1-x)}{x^3}

, is

View answer and explanation

Correct option: C

-1 / 2

Option C (-1/2) is correct because the limit is evaluated using the Taylor series expansions of

\sin x

\cos x

, and

\log_e(1-x)

330 Easy

330 of 930

Analyzing +4 | -1

\lim _{x \rightarrow+\infty} \frac{20+2 \sqrt{x}+3 x^{1 / 3}}{2+\sqrt{4 x-3}+(8 x-4)^{1 / 3}}=

View answer and explanation

Correct option: A

Option A (1) is correct because after dividing both numerator and denominator by √x, the limit as x approaches infinity is 1.

Revision notes and key takeaways

Review LIMITS in your textbook or class notes, then use this set to check your understanding.

Showing 301-330 of 930 questions

Your Progress

Limits practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice