The value of $\mathop {\lim }\limits_{x \to 3} \frac{{{3^x} - {x^2}}}{{{x^x} - {3^2}}}$, is

$\frac{{\log 3 - 1}}{{\log 3 + 1}}$

$\frac{{\log 3 + 1}}{{\log 3 - 1}}$

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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 601-630 of 930 questions

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601 Easy

601 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \pi / 6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}

equals

View answer and explanation

Correct option: B

\frac{1}{\sqrt{3}}

Option B (

\frac{1}{\sqrt{3}}

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

602 Easy

602 of 930

Analyzing +4 | -1

\alpha

is a repeated root of

a x^2+b x+c=0

, then

\lim \limits_{{x \rightarrow \alpha}} \frac{\sin \left(a x^2+b x+c\right)}{(x-\alpha)^2}

is equal to

View answer and explanation

Correct option: B

a

Option A (

a

) is correct because by factoring the quadratic and using the limit

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

, the expression simplifies to

a

603 Easy

603 of 930

Application +4 | -1

Let [ ] represents the greatest integer function less than or equal to

x

. The value of

\lim \limits_{{x \rightarrow 0}}\left(\left[\frac{n \sin x}{x}\right]+\left[\frac{n \tan x}{x}\right]\right)

View answer and explanation

Correct option: D

2 n-1

Option D (

2 n-1

) is correct because as x approaches 0,

\frac{n \sin x}{x}

approaches n from below and

\frac{n \tan x}{x}

approaches n from above, leading to the greatest integer function values of n-1 and n respectively.

604 Easy

604 of 930

Application +4 | -1

{a_1} = 1

and

{a_{n + 1}} = \frac{{4 + 3{a_n}}}{{3 + 2{a_n}}},\;n \ge 1

and if

\lim\limits_{n \to \infty } {a_n} = a

, then the value of

a

View answer and explanation

Correct option: A

\sqrt 2

Option A (

\sqrt 2

) is correct because the limit of the sequence is found by setting

a_{n+1} = a_n = a

and solving the resulting quadratic equation, discarding the negative root since all terms are positive.

605 Easy

605 of 930

Application +4 | -1

\lim _{x \rightarrow \frac{\pi}{6}}\left(\frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}\right)=

View answer and explanation

Correct option: C

-3

Option C (-3) is correct because by factoring the numerator and denominator and simplifying, the limit evaluates to -3.

606 Easy

606 of 930

Application +4 | -1

If the function

f(x)= \begin{cases}2 x & \text { for }|x| \leq 1 \\ x^2+a x+b & \text { for }|x| > 1\end{cases}

is continuous, then

View answer and explanation

Correct option: C

a=2, b=-1

Option C (

a=2, b=-1

) is correct because for the function to be continuous, the left-hand limit and right-hand limit must be equal at the points where the function definition changes, which leads to these values.

607 Easy

607 of 930

Analyzing +4 | -1

\lim\limits_{x \to {0^ - }} \frac{{\sin x}}{{\sqrt x }}

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit does not exist in the real number system due to the square root of a negative number.

608 Medium

608 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{x} = 1

, then

a,\;b

are

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given condition leads to a = b, and none of the provided options satisfy this condition.

609 Medium

609 of 930

Analyzing +4 | -1

l = \lim\limits_{x \to - 2} \frac{{\tan \pi \mathrm{ }x}}{{x + 2}} + \lim\limits_{x \to \infty } {\left( {1 + \frac{1}{{{x^2}}}} \right)^x}

then which one of the following is not correct?

View answer and explanation

Correct option: C

l < 4

Option

l < 4

is not correct because the value of

l

\pi + 1

, which is greater than 4.

610 Easy

610 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{2^{|x|} e^{|x|}-|x|-|x| \log _e 2-1}{x \tan x}

is equal to

View answer and explanation

Correct option: D

\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}

Option D (

\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}

) is correct because it is obtained by using the Taylor series expansion of

a^{|x|}

and simplifying the limit.

611 Easy

611 of 930

Analyzing +4 | -1

\operatorname{lt}_{x \rightarrow \infty}\left(\frac{x^n}{e^x}+\left(\cos \frac{1}{x}\right)^{x^2}\right)

is finite and is equal to

\mathrm{k}

, then which of the following is false ?

View answer and explanation

Correct option: C

k=\frac{1}{2}

Option C (

k=\frac{1}{2}

) is false because the limit evaluates to

e^{-\frac{1}{2}}

, thus

k = e^{-\frac{1}{2}}

612 Easy

612 of 930

Application +4 | -1

\lim \limits_ {x \rightarrow 9^{-}}\left[x^2+1\right]+\left[x^2-1\right]+\{x\}

[Here [. ] is GIF and

\{

\}

is Fractional part function]

View answer and explanation

Correct option: C

161

Option C (161) is correct because the limit of the expression as x approaches 9 from the left is 2[81] + 9 - 8 = 160 + 1 = 161.

613 Easy

613 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\cos 4 x+a \cos 2 x+b}{x^4}

is finite, then

(a, b)=

View answer and explanation

Correct option: C

(-4,3)

Option C ((-4,3)) is correct because the limit is finite only when

a = -4

and

b = 3

, which are found by applying L'Hospital's rule and setting up equations.

614 Easy

614 of 930

Analyzing +4 | -1

\lim\limits_{n \to \infty } \left( {\frac{{{1^2}}}{{1 - {n^3}}} + \frac{{{2^2}}}{{1 - {n^3}}} + ... + \frac{{{n^2}}}{{1 - {n^3}}}} \right)

is equal to

View answer and explanation

Correct option: D

- \frac{1}{6}

Option D (-1/6) is correct because after summing the series and taking the limit, the expression evaluates to -1/6.

615 Easy

615 of 930

Analyzing +4 | -1

S_n=\left(1-\tan ^4 \frac{\pi}{2^3}\right)\left(1-\tan ^4 \frac{\pi}{2^4}\right) \ldots\left(1-\tan ^4 \frac{\pi}{2^n}\right) .

The value of

\lim \limits_{{n \rightarrow \infty}} S_n

, is

View answer and explanation

Correct option: C

\frac{\pi^3}{32}

Option C (

\frac{\pi^3}{32}

) is correct because the limit of the given product is evaluated using trigonometric identities and limit properties.

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616 Easy

616 of 930

Application +4 | -1

S_1=\Sigma n, S_2=\Sigma n^2, S_3=\Sigma n^3

, then the value of

\lim \limits_{{n \rightarrow \infty}} \frac{S_1\left(1+\frac{S_3}{8}\right)}{{S_2}^2}

is equal to

View answer and explanation

Correct option: D

\frac{9}{64}

Option D (

\frac{9}{64}

) is correct because after substituting the summation formulas and simplifying, the limit evaluates to

\frac{9}{64}

617 Easy

617 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \pi}} \frac{\sin 3 x}{\sin 2 x}

is equal to

View answer and explanation

Correct option: D

-\frac{3}{2}

The correct answer is D (

-\frac{3}{2}

) because it is the limit calculated using L'Hopital's rule or trigonometric identities.

618 Easy

618 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0}\left(\frac{1-2^x-5^x+10^x}{x \sin x}\right)=

View answer and explanation

Correct option: C

\log _e 5 \cdot \log _e 2

Option C (

\log _e 5 \cdot \log _e 2

) is correct because the limit can be simplified by factoring and using standard limit formulas.

619 Easy

619 of 930

Analyzing +4 | -1

\lim _{n \rightarrow \infty}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0

, then

8(\alpha+\beta)

is equal to

View answer and explanation

Correct option: C

-4

Option C (-4) is correct because, after rationalizing and comparing coefficients,

8(\alpha + \beta)

equals -4.

620 Medium

620 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} n \cos \left(\frac{\pi}{4 n}\right) \sin \left(\frac{\pi}{4 n}\right)=k

, then

k

View answer and explanation

Correct option: A

\frac{\pi}{4}

Option A (

\frac{{\pi}}{{4}}

) is correct because applying the limit and using the small angle approximation yields

\frac{{\pi}}{{4}}

621 Medium

621 of 930

Analyzing +4 | -1

\lim\limits_{x \to \infty } \left( {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right)

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (1/2) is correct because after rationalizing and simplifying, the limit as x approaches infinity evaluates to 1/2.

622 Easy

622 of 930

Analyzing +4 | -1

Let

f(x)=[x]+[-x]

, where

[x]

denotes the greatest integer less than or equal to

x

. Then, for any integer

m

View answer and explanation

Correct option: B

\lim \limits_{{x \rightarrow m}} f(x) \neq f(m)

Option B (

\lim \limits_{{x \rightarrow m}} f(x) \neq f(m)

) is correct because the limit as x approaches m is -1, while f(m) = 0.

623 Easy

623 of 930

Application +4 | -1

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

is equal to:

View answer and explanation

Correct option: C

e^{-5}

Option C (e^-5) is correct because the limit is of the form 1^infinity, which can be solved using the exponential limit formula.

624 Easy

624 of 930

Application +4 | -1

For

x \in R, \lim \limits_{{x \rightarrow \infty}}\left(\frac{x-3}{x+2}\right)^x=

View answer and explanation

Correct option: C

e^{-5}

Option C,

e^{-5}

, is correct because it is the result of rewriting the limit in the form of

e^{\lim x}

and then evaluating the limit.

625 Easy

625 of 930

Analyzing +4 | -1

f(x)= \begin{cases}\frac{\sin [x]}{[x]}, & {[x] \neq 0,} \\ 0, & \text [x] \neq 0,\end{cases}

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

None of these

The correct answer is D (None of these) because the limit does not exist as the left-hand limit and right-hand limit are not equal.

626 Easy

626 of 930

Analyzing +4 | -1

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

equals

View answer and explanation

Correct option: A

4 \sqrt{2}

Option A (

4 \sqrt{2}

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

4 \sqrt{2}

627 Easy

627 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 3} \frac{{{3^x} - {x^2}}}{{{x^x} - {3^2}}}

, is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit, after applying L'Hopital's rule, evaluates to a different expression than the given options.

628 Medium

628 of 930

Analyzing +4 | -1

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

View answer and explanation

Correct option: B

{e^{2/\pi }}

Option B (

{e^{2/\pi }}

) is correct because the limit can be evaluated by substituting and using standard limit formulas.

629 Easy

629 of 930

Understanding +4 | -1

The value of

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

View answer and explanation

Correct option: A

Option A (0) is correct because as x approaches 1 from the left,

\sin(\sin^{-1}x)

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

630 Medium

630 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\frac{x+7}{x+2}\right)^{x+4}

View answer and explanation

Correct option: D

e^5

Option D (

e^5

) is correct because it's the result of evaluating the limit using the standard limit formula for exponential functions.

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