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 $\left| x \right| = 1$

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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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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.

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Showing 181-210 of 647 questions

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181 Easy

181 of 647

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\log \left( {1 + {x^3}} \right)}}{{{{\sin }^3}x}}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because by using the series expansion of log(1+x) and sin(x), the limit simplifies to 1.

182 Easy

182 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow \pi / 3}} \frac{2 \sin (x-\pi / 3)}{1-2 \cos x}

View answer and explanation

Correct option: B

2 / \sqrt{3}

Option B (

2 / \sqrt{3}

) is correct because L'Hopital's rule is applied to evaluate the limit of the given expression as x approaches

\pi/3

183 Easy

183 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}}\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right)

, is

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (2/3) is correct because the limit is evaluated by rationalizing the expression and simplifying.

184 Easy

184 of 647

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: A

-\frac{1}{2}

Option A (-1/2) is correct because it is obtained by applying the limit properties and using the known limit

\lim \limits_{{x \rightarrow 0}} \frac{{\tan x-\sin x}}{{\sin ^3 x}}=\frac{{1}}{{2}}

185 Easy

185 of 647

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(2^n+5^n\right)^{1 / n}

is equal to

View answer and explanation

Correct option: B

Option B, 5, is correct because it is the result of factoring out the largest term and then taking the limit.

186 Easy

186 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 1}} \sec \frac{\pi}{2 x} \log x

equals

View answer and explanation

Correct option: C

\frac{2}{\pi}

Option C (

\frac{2}{\pi}

) is correct because after applying L'Hôpital's rule, the limit evaluates to

\frac{2}{\pi}

187 Easy

187 of 647

Analyzing +4 | -1

The value of

\lim\limits_{n \to \infty } \left\{ {\frac{1}{3} + \frac{2}{{21}} + \frac{3}{{91}} + \ldots + \frac{n}{{{n^4} + {n^2} + 1}}} \right\}

, is

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because the limit of the given series as n approaches infinity is 1/2, obtained by telescoping the sum.

188 Easy

188 of 647

Application +4 | -1

\lim\limits_{x \to \infty } x\sin \left( {\frac{2}{x}} \right)

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because by substituting

u = \frac{2}{x}

, the limit transforms into a standard limit:

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

189 Easy

189 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{x^8-2 x+1}{x^4-2 x+1}

equals

View answer and explanation

Correct option: A

Option A (3) is correct because it is the result of factoring and simplifying the limit expression as x approaches 1.

190 Easy

190 of 647

Application +4 | -1

Let

{f}({x})= \begin{cases}{x}+2, & {x} \leq-1 \\ {cx}^2, & {x}>-1\end{cases}

\lim \limits_{{x \rightarrow-1}} f(x)

exists, then

c

is equal to

View answer and explanation

Correct option: A

Option A is correct because for the limit to exist at x = -1, the value of c must be 1.

191 Easy

191 of 647

Analyzing +4 | -1

f(x)=\left\{\begin{array}{cc}\frac{\sin [x]}{[x]}, & \text { for }[x] \neq 0 \\ 0, & \text { for }[x]=0\end{array}\right.

, where

[x]

denotes the greatest integer less than or equal to

x

, then

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

equals

View answer and explanation

Correct option: D

doesn't exist

Option D (doesn't exist) is correct because the left-hand limit and the right-hand limit at x=0 are not equal, indicating that the limit does not exist.

192 Easy

192 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}}(1+\sin x)^{2 \cot x}

, is

View answer and explanation

Correct option: B

e^2

Option B (

e^2

) is correct because the limit can be evaluated using the exponential form of limits and trigonometric identities.

193 Easy

193 of 647

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}

194 Easy

194 of 647

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}

195 Easy

195 of 647

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.

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196 Easy

196 of 647

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

197 Easy

197 of 647

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.

198 Easy

198 of 647

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.

199 Easy

199 of 647

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}

200 Easy

200 of 647

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.

201 Easy

201 of 647

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.

202 Easy

202 of 647

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.

203 Easy

203 of 647

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

204 Easy

204 of 647

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}

205 Easy

205 of 647

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.

206 Easy

206 of 647

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.

207 Easy

207 of 647

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}

208 Easy

208 of 647

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.

209 Easy

209 of 647

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.

210 Easy

210 of 647

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.

Revision notes and key takeaways

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