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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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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 901-930 of 930 questions

901 Medium

901 of 930

Analyzing +4 | -1

\lim\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}

is given by

View answer and explanation

Correct option: B

\frac{1}{{\sqrt {2\pi } }}

Option B (

\frac{1}{{\sqrt {2\pi } }}

) is correct because by substituting

{\cos ^{ - 1}}x = y

and applying trigonometric identities and limit properties, the expression simplifies to

\frac{1}{{\sqrt {2\pi } }}

902 Easy

902 of 930

Application +4 | -1

f(3)=6

f^{\prime}(3)=2

, then

\operatorname{Limit}_{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}

is given by :

View answer and explanation

Correct option: C

Option C (0) is correct because when applying L'Hopital's rule and using the given values of f(3) and f'(3), the limit evaluates to zero.

903 Easy

903 of 930

Application +4 | -1

Let

f(x)= \begin{cases}|x-3| & \text { if } x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4} & \text { if } x<1\end{cases}

Then

\lim _{x \rightarrow 1} f(x)

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because the left-hand limit and right-hand limit at x=1 both equal 2, hence the limit exists and is equal to 2.

904 Easy

904 of 930

Application +4 | -1

\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)

View answer and explanation

Correct option: A

The correct answer is A (0) because simplifying the expression and applying L'Hôpital's rule yields the value 0.

905 Easy

905 of 930

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin \{ \cos x\} }}{{x - \frac{\pi }{2}}},}\,{x \ne \frac{\pi }{2}} \\ {1,}\,{x = \frac{\pi }{2}} \end{array}} \right.

, where

\{.\}

represents the fractional part function, then

\lim \limits_{{x \rightarrow \pi / 2}} f(x)

, is

View answer and explanation

Correct option: C

Non-existent

Option C (Non-existent) is correct because the left-hand limit and right-hand limit are not equal, hence the limit does not exist.

906 Easy

906 of 930

Application +4 | -1

3-\frac{x^2}{12} \leq f(x) \leq 3+\frac{x^3}{9}

for all

x \neq 0

, then the value of

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

is equal to

View answer and explanation

Correct option: B

Option B (3) is correct because, according to the Sandwich Theorem, the limit of f(x) as x approaches 0 is squeezed between two functions that both approach 3.

907 Easy

907 of 930

Application +4 | -1

Let

f(x)=\left\{\begin{array}{ccc}a x-b, & x \leq 1 \\ 3 x, & 1 < x < 2 \\ b x^2-a, & x \geq 2\end{array}\right.

f

is continuous function then

(a, b)

is equal to

View answer and explanation

Correct option: D

(6,3)

Option D ((6,3)) is correct because for f to be continuous, the left and right limits at x=1 and x=2 must be equal, leading to the solution a=6 and b=3.

908 Easy

908 of 930

Application +4 | -1

\lim\limits_{x \to a} \left( {\frac{{\tan x - \tan a}}{{x - a}}} \right)

View answer and explanation

Correct option: D

{\sec ^2}a

Option D (

{\sec ^2}a

) is correct because the limit represents the derivative of the tangent function evaluated at x=a, and the derivative of tan(x) is sec²(x).

909 Easy

909 of 930

Analyzing +4 | -1

0 < x < y

, then

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

is equal to

View answer and explanation

Correct option: C

y

Option C (

y

) is correct because as

n

approaches infinity, the term with the larger base (

y

) dominates the expression.

910 Medium

910 of 930

Analyzing +4 | -1

f(x)=x \frac{e^{[x]+|x|}-2}{[x]+|x|}

, then

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

, is

View answer and explanation

Correct option: D

Non-existent

Option D (Non-existent) is correct because the left-hand limit and right-hand limit are not equal, so the limit does not exist.

911 Easy

911 of 930

Application +4 | -1

\lim _{x \rightarrow 0} \frac{\sin (a+x)-\sin (a-x)}{x}

equals

View answer and explanation

Correct option: D

2 \cos a

Option D (

2 \cos a

) is correct because using the trigonometric identity for the difference of sines simplifies the expression, and the limit of

\frac{\sin x}{x}

as x approaches 0 is 1.

912 Easy

912 of 930

Application +4 | -1

\lim \limits_{x \rightarrow a^{-}} \frac{\sqrt{x-b}-\sqrt{a-b}}{\left(x^2-a^2\right)},(a>b)

View answer and explanation

Correct option: D

\frac{1}{4 a \sqrt{a-b}}

Option D (

\frac{1}{4 a \sqrt{a-b}}

) is correct because the limit can be evaluated by rationalizing the numerator and simplifying the expression.

913 Easy

913 of 930

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

{e^4}

Option C (

{e^4}

) is correct because it is the result of evaluating the limit of the given expression as x approaches infinity, using the standard limit form for e.

914 Easy

914 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\tan x-x}{x^2 \tan x}

equals

View answer and explanation

Correct option: C

1 / 3

Option C (

1/3

) is correct because the limit is evaluated using L'Hospital's Rule multiple times to resolve the indeterminate form.

915 Medium

915 of 930

Application +4 | -1

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

View answer and explanation

Correct option: D

Option D (1) is correct because by using the small angle approximations or L'Hopital's rule, the limit simplifies to 1.

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916 Easy

916 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(1+\sin \frac{a}{n}\right)^n

equals

View answer and explanation

Correct option: A

e^a

Option A (

e^a

) is correct because the limit can be evaluated using the exponential form of limits and the standard limit

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

917 Easy

917 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{{\sin }^2}x + \cos x - 1}}{{{x^2}}}

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

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

918 Easy

918 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because after rationalizing and simplifying, the limit as x approaches infinity is

\frac{1}{2}

919 Easy

919 of 930

Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{4^{1 / n}-1}{3^{1 / n}-1}

is equal to

View answer and explanation

Correct option: B

\log _3 4

Option B (

\log_3 4

) is correct because the limit is evaluated using the standard limit

\lim_{x \to 0} \frac{a^x - 1}{x} = \log a

920 Medium

920 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{\log [x]}{x}

, where

[x]

denotes the greatest integer less than or equal to

x

, is

View answer and explanation

Correct option: A

Option A (0) is correct because the limit is evaluated using L'Hospital's Rule after applying the greatest integer function.

921 Medium

921 of 930

Analyzing +4 | -1

f(x)

is a polynomial satisfying

f(x)\cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)

and

f(2) > 1

, then

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

View answer and explanation

Correct option: A

Option A (2) is correct because the polynomial satisfying the given condition is

f(x) = x^n + 1

, and thus

\lim_{x \rightarrow 1} f(x) = 2

922 Easy

922 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} \,\,\frac{{{e^{\tan x - {e^x}}}}}{{\tan x - x}} =

View answer and explanation

Correct option: A

Option A (1) is correct because after factoring out

e^x

and using the limit property

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

, the limit simplifies to 1.

923 Easy

923 of 930

Analyzing +4 | -1

a > 0

and

\lim \limits_{{x \rightarrow \infty}}\left\{\sqrt{x^2+x+1}-(a x+b)\right\}=0

, then

(a, b)

lies on the line

View answer and explanation

Correct option: B

3 x+4 y-5=0

Option A (l=m) is correct because both limits, as x approaches infinity and negative infinity, evaluate to 1/2.

924 Easy

924 of 930

Application +4 | -1

The left hand and right hand limits of the following function

f(x)= \begin{cases}5 x-4, & 0 < x \leq 1 \\ 4 x^3-3 x, & 1 < x < 2\end{cases}

x=1

, are

View answer and explanation

Correct option: A

1, 1

Option A (1, 1) is correct because both the left-hand limit and the right-hand limit of the function at x=1 are equal to 1.

925 Easy

925 of 930

Analyzing +4 | -1

\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} \text { is: }

View answer and explanation

Correct option: D

Option D (8) is correct because after applying trigonometric identities and simplifying, the limit evaluates to 8.

926 Easy

926 of 930

Analyzing +4 | -1

f(x)=\frac{1}{3}\left(f(x+1)+\frac{5}{f(x+2)}\right)

and

f(x)>0, \forall x \in R

, then

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

View answer and explanation

Correct option: C

\sqrt{\frac{5}{2}}

The correct answer is C (

\sqrt{\frac{{5}}{{2}}}

) because by assuming the limit exists and solving for it, we find that the limit is

\sqrt{\frac{{5}}{{2}}}

927 Easy

927 of 930

Analyzing +4 | -1

|x| < 1

, then

\lim \limits_{{n \rightarrow \infty}}\left\{(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots .\left(1+x^{2 n}\right)\right\}

is equal to

View answer and explanation

Correct option: B

\frac{1}{1-x}

Option B (

\frac{{1}}{{1-x}}

) is correct because the infinite product can be simplified to a telescoping product, and the limit as n approaches infinity is

\frac{{1}}{{1-x}}

928 Easy

928 of 930

Analyzing +4 | -1

The value of the limit

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

is equal to

View answer and explanation

Correct option: A

- \frac{1}{2}

Option A (-1/2) is correct. The limit can be evaluated using trigonometric identities and L'Hopital's rule or by converting to sine functions and using limit properties.

929 Easy

929 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{x^{2^{32}}-2^{32} x+4^{16}-1}{(x-1)^2}

is equal to

View answer and explanation

Correct option: A

2^{63}-2^{31}

Option A (

2^{63}-2^{31}

) is correct because the limit is evaluated by applying L'Hopital's rule twice and simplifying the expression.

930 Easy

930 of 930

Analyzing +4 | -1

x > 0

and

g

is a bounded function, then

\lim \limits_{{n \rightarrow \infty}} \frac{f(x) e^{n x}+g(x)}{e^{n x}+1}

View answer and explanation

Correct option: B

f(x)

Option B (

f(x)

) is correct because as n approaches infinity, the expression simplifies to

f(x)

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