If $a,b,c,d$ are positive, then $\mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{1}{{a + bx}}} \right)^{c + dx}} = $

${e^{\left( {c + d} \right)/a + b}}$

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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 301-330 of 647 questions

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301 Easy

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

302 Easy

302 of 647

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.

303 Easy

303 of 647

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

304 Easy

304 of 647

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.

305 Easy

305 of 647

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 }}

306 Easy

306 of 647

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}

307 Easy

307 of 647

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.

308 Easy

308 of 647

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.

309 Easy

309 of 647

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.

310 Easy

310 of 647

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.

311 Easy

311 of 647

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).

312 Easy

312 of 647

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.

313 Easy

313 of 647

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}

314 Easy

314 of 647

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.

315 Easy

315 of 647

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.

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316 Easy

316 of 647

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.

317 Easy

317 of 647

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.

318 Easy

318 of 647

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.

319 Easy

319 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{\cot ^{-1}\left(x^{-a} \log _a x\right)}{\sec ^{-1}\left(a^x \log _x a\right)}(a>1)

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because as x approaches infinity, both the numerator and denominator approach

\frac{\pi}{2}

, resulting in a limit of 1.

320 Easy

320 of 647

Application +4 | -1

f\left( x \right) = {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 2}}} \right)^x}

then

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

is equal to

View answer and explanation

Correct option: A

{e^4}

Option A (

{e^4}

) is correct because by rewriting the expression and taking the limit as x approaches infinity, the result is

{e^4}

321 Easy

321 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\cos (\sin x)-1}{x^2}=

View answer and explanation

Correct option: D

-1 / 2

Option D (-1/2) is correct because the limit can be evaluated using the standard limit

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

and

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

322 Easy

322 of 647

Analyzing +4 | -1

\lim\limits_{x \to \frac{\pi }{2}}\left(\tan ^2 x\left(\left(2 \sin ^2 x+3 \sin x+4\right)^{\frac{1}{2}}-\left(\sin ^2 x+6 \sin x+2\right)^{\frac{1}{2}}\right)\right)

is equal to

View answer and explanation

Correct option: A

\frac{1}{12}

Option A (

\frac{1}{12}

) is correct because after rationalizing and applying L'Hopital's rule or trigonometric identities, the limit evaluates to

\frac{1}{12}

323 Easy

323 of 647

Understanding +4 | -1

The value of

\lim \limits_{x \rightarrow 0^{+}}\left[\frac{\sin x}{x}\right]

View answer and explanation

Correct option: B

Option B (0) is correct because as x approaches 0 from the right,

\frac{{\sin x}}{{x}}

approaches 1 from below, and the greatest integer function of a value approaching 1 from below is 0.

324 Easy

324 of 647

Application +4 | -1

a,b,c,d

are positive, then

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

View answer and explanation

Correct option: A

{e^{d/b}}

Option A (

{e^{d/b}}

) is correct because the limit can be rewritten in the form of

e

raised to the power of another limit, which simplifies to

{e^{d/b}}

325 Easy

325 of 647

Application +4 | -1

\lim\limits_{n \to \infty } \frac{{1 - {{\left( {10} \right)}^n}}}{{1 + {{\left( {10} \right)}^{n + 1}}}} = \frac{{ - \alpha }}{{10}}

, then the value of

\alpha

View answer and explanation

Correct option: C

Option C (1) is correct because it is the result of dividing both the numerator and denominator by

10^n

and then evaluating the limit as n approaches infinity.

326 Easy

326 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because by factoring out

e^x

and applying the standard limit, the expression simplifies to 1.

327 Easy

327 of 647

Application +4 | -1

{l_1} = \lim\limits_{x \to {2^ + }} \left( {x + \left[ x \right]} \right),{l_2} = \lim\limits_{x \to {2^ - }} \left( {2x - \left[ x \right]} \right)

and

{l_3} = \lim\limits_{x \to \pi /2} \frac{{\cos x}}{{\left( {x - \pi /2} \right)}}

, then

View answer and explanation

Correct option: C

{l_3} < {l_2} < {l_1}

Option C (

{l_3} < {l_2} < {l_1}

) is correct because evaluating each limit separately gives l1 = 4, l2 = 3, and l3 = -1, which satisfy the inequality.

328 Easy

328 of 647

Analyzing +4 | -1

\lim\limits_{x \to \infty } \left\{ {ax - \frac{{{x^2} + 1}}{{x + 1}}} \right\} = b,\;

, a finite number, then

View answer and explanation

Correct option: A

a = 1,\;b = 1

Option A (

a = 1, b = 1

) is correct because for the limit to be finite, the

x^2

term must vanish, leading to

a = 1

, and then the limit evaluates to

b = 1

329 Easy

329 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} x^8\left[\frac{1}{x^3}\right]

, where [.] denotes the greatest integer function is

View answer and explanation

Correct option: C

Option C (0) is correct because by using the squeeze theorem, we can show that the limit of the given expression as x approaches 0 is 0.

330 Easy

330 of 647

Application +4 | -1

Let

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

for

n=1,2,3, \ldots

. Then the value of

\lim _{n \rightarrow \infty} x_n

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

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

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

, and dividing by

n^2

and taking the limit as

n

approaches infinity yields

\frac{1}{2}

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