If ${A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\;i = 1,\;2,\; \ldots ,\;n$ and if ${a_1} < {a_2} < {a_3} < \ldots < {a_n}$. Then, $\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2} \ldots {A_n}} \right),\;1 \le m \le n$

Is equal to ${\left( { - 1} \right)^m}$

Is equal to ${\left( { - 1} \right)^{m + 1}}$

Is equal to ${\left( { - 1} \right)^{m - 1}}$

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

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

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 481-510 of 930 questions

Preparing practice and saved progress…

481 Medium

481 of 930

Analyzing +4 | -1

{A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\;i = 1,\;2,\; \ldots ,\;n

and if

{a_1} < {a_2} < {a_3} < \ldots < {a_n}

. Then,

\lim\limits_{x \to {a_m}} \left( {{A_1}{A_2} \ldots {A_n}} \right),\;1 \le m \le n

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left-hand limit and right-hand limit of the product of A_i's as x approaches a_m are not equal, thus the limit does not exist.

482 Easy

482 of 930

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.

483 Easy

483 of 930

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}

484 Easy

484 of 930

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

485 Easy

485 of 930

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}

486 Easy

486 of 930

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.

487 Easy

487 of 930

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}}

488 Easy

488 of 930

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.

489 Medium

489 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 2} \frac{5}{{\sqrt 2 - \sqrt x }},

View answer and explanation

Correct option: D

Non-existent

Option D (Non-existent) is correct because the left-hand limit approaches infinity and the right-hand limit approaches negative infinity.

490 Easy

490 of 930

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.

491 Easy

491 of 930

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.

492 Easy

492 of 930

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

493 Easy

493 of 930

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.

494 Easy

494 of 930

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}

495 Easy

495 of 930

Analyzing +4 | -1

Let

f, g

and

h

be the real valued functions defined of

\mathbb{R}

f(x)\left\{\begin{array}{ll} \frac{x}{|x|}, & x \neq 0 \\ 1, & x=0 \end{array}, \,\,\, g(x) \begin{cases}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{cases}\right.

and

h(x)=2[x]-f(x)

, where

[x]

is the greatest integer

\leq x

.

Then the value of

\lim \limits_{{x \rightarrow 1}} g(h(x-1))

View answer and explanation

Correct option: A

Option A is correct because the limit of

g(h(x-1))

x

approaches 1 is 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

496 Hard

496 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow-1}} \frac{\cos 2-\cos 2 x}{x^2-|x|}

is equal to

View answer and explanation

Correct option: C

2 \sin 2

Option C (

2 \sin 2

) is correct because the limit is evaluated using trigonometric identities and limit properties, resulting in

2 \sin 2

497 Medium

497 of 930

Analyzing +4 | -1

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}

equals :

View answer and explanation

Correct option: C

\frac{1}{16}

Option A (

-\sin 1

) is correct because after substituting and simplifying, the limit equals

-\sin 1

498 Medium

498 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

-\frac{1}{2}

Option C (

-\frac{1}{2}

) is correct because after applying the series expansions of sin(x) and log(1-x), the limit evaluates to -1/2.

499 Medium

499 of 930

Analyzing +4 | -1

\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because applying L'Hopital's rule to the limit results in the expression

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

500 Medium

500 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} 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 different, hence the limit does not exist.

501 Medium

501 of 930

Application +4 | -1

f^{\prime}(2)=1

, then

\lim \limits_{{h \rightarrow 0}} \frac{f\left(2+h^2\right)-f\left(2-h^2\right)}{2 h^2}=\ldots \ldots

View answer and explanation

Correct option: B

Option B (1) is correct because by applying L'Hopital's rule, the limit simplifies to f'(2), which is given as 1.

502 Easy

502 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{\sin ^2\left(x^3+x^2+x-3\right)}{1-\cos \left(x^2-4 x+3\right)}

has the value equal to

View answer and explanation

Correct option: A

Option A (18) is correct because after applying L'Hopital's rule and simplifying, the limit evaluates to 18.

503 Easy

503 of 930

Understanding +4 | -1

f(x)=|x|-5

, then the value of

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

View answer and explanation

Correct option: A

Option A (0) is correct because as x approaches 5, |x| approaches 5, and |x| - 5 approaches 0.

504 Easy

504 of 930

Application +4 | -1

G(x)=-\sqrt{25-x^2}

, then

\lim _{x \rightarrow 1}\left(\frac{G(x)-G(1)}{x-1}\right)=

View answer and explanation

Correct option: A

\frac{1}{2 \sqrt{6}}

Option A (

\frac{1}{2 \sqrt{6}}

) is correct because, after applying L'Hopital's rule or rationalizing, the limit of the expression as x approaches 1 is

\frac{1}{2 \sqrt{6}}

505 Easy

505 of 930

Application +4 | -1

Let

f:R \to R

be a differentiable function and

f\left( 1 \right) = 4

. Then, the value of

\lim\limits_{x \to 1} \,\,\int \nolimits_4^{f\left( x \right)} \,\,\frac{{2\,t}}{{x - 1}}\,\,dt

〗is

View answer and explanation

Correct option: A

8\,\,f'\,\,\left( 1 \right)

Option A (

8 f'(1)

) is correct because it is the result of applying L'Hopital's rule and the Fundamental Theorem of Calculus to the given limit.

506 Easy

506 of 930

Analyzing +4 | -1

0 < p < q

, then

\lim\limits_{n \to \infty } {\left( {{q^n} + {p^n}} \right)^{1/n}}

is equal to

View answer and explanation

Correct option: C

q

Option C (

q

) is correct because as

n

approaches infinity, the term with the larger base (

q

) dominates, and the limit approaches

q

507 Easy

507 of 930

Application +4 | -1

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

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because after simplification and applying the limit, the expression evaluates to 1/2.

508 Easy

508 of 930

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\sin 5x}}{{{x^2}\sin 3x}}

equals

View answer and explanation

Correct option: A

\frac{{10}}{3}

Option A (

\frac{{10}}{3}

) is correct because by applying trigonometric identities and standard limits, the expression simplifies to this value.

509 Medium

509 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right) \ldots\left(1+\frac{1}{a_n}\right)

, where

a_1=1

and

a_n=n\left(1+a_{n-1}\right), \forall n \geq 2

, is

View answer and explanation

Correct option: A

e

Option A (e) is correct because the given limit simplifies to the series expansion of e after using the recurrence relation and taking the limit as n approaches infinity.

510 Easy

510 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{\left[ {\left( {a - n} \right)nx - \tan x} \right]\sin nx}}{{{x^2}}} = 0

, where

n

is non-zero real number, then a is equal to

View answer and explanation

Correct option: D

n + \frac{1}{n}

Option D (

n + \frac{1}{n}

) is correct because substituting the limit into the given equation and solving for 'a' yields this result.

Revision notes and key takeaways

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

Showing 481-510 of 930 questions

Your Progress

Limits practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice