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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 721-750 of 930 questions

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721 Easy

721 of 930

Application +4 | -1

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

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because the limit can be evaluated by rewriting the expression and using the standard limit definition of

e

722 Easy

722 of 930

Understanding +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{a x^2+b x+c}{d x^2+e x+f}

, is

View answer and explanation

Correct option: A

\frac{a}{d}

Option A (

\frac{a}{d}

) is correct because when x approaches infinity, the highest power terms dominate the limit.

723 Easy

723 of 930

Application +4 | -1

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

then

\lim _{n \rightarrow \infty} y_n

is equal to

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

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

\frac{n(n+1)(2n+1)}{6}

, and dividing by

n^3

and taking the limit as

n

approaches infinity yields

\frac{1}{3}

724 Easy

724 of 930

Application +4 | -1

For

x \in R\,\,\lim\limits_{x \to \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 by rewriting the expression and using the limit definition of

e

, the limit evaluates to

{e^{ - 5}}

725 Easy

725 of 930

Application +4 | -1

\lim\limits_{x \to 1} \left( {1 + \cos \pi \mathrm{ }x} \right){\cot ^2}\pi

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 1/2.

726 Medium

726 of 930

Application +4 | -1

Let

f\left( 2 \right) = 4

and

f'\left( 2 \right) = 4

. Then,

\lim\limits_{x \to 2} \frac{{xf\left( 2 \right) - 2f\left( x \right)}}{{x - 2}}

is given by

View answer and explanation

Correct option: C

- 4

Option C (

-4

) is correct because it is obtained by applying L'Hopital's rule or by rewriting the limit in terms of the derivative of f(x).

727 Medium

727 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} {\left( {{\text{cosec}}\,\,x} \right)^{\frac{1}{{{{\log }_e}x}}}}

equals

View answer and explanation

Correct option: C

\frac{1}{e}

Option C (

\frac{1}{e}

) is correct because by applying L'Hopital's rule to the logarithm of the expression, we find the limit is -1, thus the original limit is

\frac{1}{e}

728 Easy

728 of 930

Analyzing +4 | -1

The value of

\lim _{x \rightarrow 0^{+}} x^m(\log x)^n, m

\mathrm{n} \in \mathrm{N}

View answer and explanation

Correct option: A

Option A is correct. By applying L'Hopital's rule repeatedly, the limit can be evaluated to 0.

729 Easy

729 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} k x \operatorname{cosec} x=\lim \limits_{{x \rightarrow 0}} x \operatorname{cosec} k x

, then

k=

View answer and explanation

Correct option: C

\pm 1

Option C (

\pm 1

) is correct because solving the equation

\lim \limits_{{x \rightarrow 0}} k x \operatorname{cosec} x=\lim \limits_{{x \rightarrow 0}} x \operatorname{cosec} k x

yields

k = \pm 1

730 Medium

730 of 930

Application +4 | -1

\lim \limits_ {\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=\ldots \ldots

View answer and explanation

Correct option: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct because the limit can be evaluated using L'Hopital's rule, resulting in cos(

\alpha

) + sin(

\alpha

), which evaluates to

\sqrt{2}

\alpha

\pi

/4.

731 Medium

731 of 930

Analyzing +4 | -1

Let

f(x)=\sin x / x

for

x \neq 0

and [.] denote the greatest integer function. Then

\lim _{x \rightarrow 0}[f(x)]

View answer and explanation

Correct option: A

Option A (0) is correct because for x near 0, but not equal to 0,

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

is between 0 and 1, so its greatest integer function is 0.

732 Medium

732 of 930

Application +4 | -1

Let

f:R \to R

be a differentiable function having

f\left( 2 \right) = 6,\;f'\left( 2 \right) = \left( {\frac{1}{{48}}} \right)

. Then,

\lim\limits_{x \to 2} \frac{{\int \nolimits_6^{f\left( x \right)} 4\;{t^3}\;dt}}{{x - 2}}

is equals

View answer and explanation

Correct option: A

Option A (18) is correct because it is the result of applying L'Hopital's rule and the Fundamental Theorem of Calculus to the given limit.

733 Medium

733 of 930

Analyzing +4 | -1

f(x+y)=f(x)+f(y)

for all

x, y \in R

and

f(1)=1

, then

\lim \limits_{{x \rightarrow 0}} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^2 f(\sin x)}

equals

View answer and explanation

Correct option: C

\log \sqrt{2}

Option C (

\log \sqrt{2}

) is correct because after simplifying the expression using the given functional equation and limit properties, the limit evaluates to

\frac{1}{2} \log 2 = \log \sqrt{2}

734 Easy

734 of 930

Analyzing +4 | -1

x > 0

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

View answer and explanation

Correct option: C

Option C (

\sin 1

) is correct. The limit is found by evaluating the left-hand limit and right-hand limit separately and showing that they are equal to

\sin 1

735 Easy

735 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^n \cot ^{-1}\left(\frac{r^3-r+\frac{1}{r}}{2}\right)

View answer and explanation

Correct option: C

\frac{\pi}{2}

Option C (

\frac{\pi}{2}

) is correct because the given sum telescopes after applying the arctangent subtraction formula, and the limit as n approaches infinity is

\frac{\pi}{2}

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736 Easy

736 of 930

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {{e^{kx}} - 1} \right)\sin kx}}{{{x^2}}} = 4

, then

k

is equal to

View answer and explanation

Correct option: C

\pm \,2

Option C (

\pm \,2

) is correct because solving the limit equation yields two possible values for k, both positive and negative 2.

737 Easy

737 of 930

Application +4 | -1

The value of

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

is:

View answer and explanation

Correct option: B

e^a

Option B (

e^a

) is correct because the limit can be evaluated using the standard limit formula for exponential functions and L'Hopital's rule.

738 Easy

738 of 930

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{x + 6}}{{x + 1}}} \right)^{x + 4}}

, is

View answer and explanation

Correct option: D

{e^5}

Option D (

{e^5}

) 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.

739 Easy

739 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}|x|^{[\cos x]}

is equal to

View answer and explanation

Correct option: A

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

\cos x

approaches 1, and any number (except 0) raised to the power of 0 is 1.

740 Medium

740 of 930

Analyzing +4 | -1

a.\lim \limits_{{x \rightarrow 1}} x^{1 / 1-x}+b=e^{-1}(a \geq 1, b \geq 0)

, then

View answer and explanation

Correct option: D

a=1, b=0

Option D (

a=1, b=0

) is correct because by evaluating the limit and comparing with the given equation, we find a=1 and b=0.

741 Easy

741 of 930

Application +4 | -1

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

View answer and explanation

Correct option: C

3 / 2

Option C (

3 / 2

) is correct because the limit is evaluated by using trigonometric identities and standard limit results.

742 Easy

742 of 930

Analyzing +4 | -1

Let

f(x)

be a real valued function defined for all

x \geq 1

, satisfying

f(1)=1

and

f^{\prime}(x)=\frac{1}{x^2+(f(x))}

; then

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

View answer and explanation

Correct option: C

exists and less than

1+\frac{\pi}{4}

Option C, exists and less than

1 + \frac{\pi}{4}

, is correct because

f'(x) > 0

implies

f(x)

is increasing, and by comparing

f'(t)

with

\frac{1}{1+t^2}

and integrating, we find that

\lim_{x \to \infty} f(x) < 1 + \frac{\pi}{4}

743 Medium

743 of 930

Analyzing +4 | -1

\lim\limits _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}

, 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 as x approaches 0.

744 Easy

744 of 930

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: D

Option D (1) is correct because, as x approaches infinity, the base approaches 1 and the exponent approaches 0, resulting in a limit of 1.

745 Medium

745 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 1} {\left( {{{\log }_2}2x} \right)^{{{\log }_x}5}}

View answer and explanation

Correct option: B

{e^{{{\log }_2}5}}

Option B (

{e^{{{\log }_2}5}}

) is correct because after applying logarithmic properties and simplifying, the limit evaluates to

{e^{{{\log }_2}5}}

746 Easy

746 of 930

Understanding +4 | -1

The value of

\lim\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)

View answer and explanation

Correct option: C

Option C (1) is correct because as x approaches infinity, the highest power terms dominate, and the ratio of the coefficients of x² in the numerator and denominator is 1.

747 Easy

747 of 930

Analyzing +4 | -1

\lim \limits_{{h \rightarrow 0}} \frac{e^x+e^{-x}-2 \cos x}{2 x \sin x}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because after applying L'Hôpital's rule twice, the limit evaluates to 1.

748 Easy

748 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{2 x^{1 / 2}+3 x^{1 / 3}+4 x^{1 / 4}+\ldots n x^{1 / n}}{(2 x-3)^{1 / 2}+(2 x-3)^{1 / 3}+\ldots+(2 x-3)^{1 / n}}

View answer and explanation

Correct option: C

\sqrt{2}

Option C (

\sqrt{2}

) is correct because it is obtained by dividing each term by

x^{1/2}

and taking the limit as x approaches infinity.

749 Hard

749 of 930

Application +4 | -1

\mathop {{\text{Lim}}}\limits_{x \to 0} \left( {1 + 2 + 3 + ........ + \left[ {\frac{1}{{\left| {\text{x}} \right|}}} \right]} \right)

where [.] denotes greatest integer function , is equal to

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct. By evaluating the limit as x approaches 0, the expression converges to

\frac{1}{2}

750 Medium

750 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}

is equal to

View answer and explanation

Correct option: B

1/6

Option B (

1/6

) is correct because the limit is evaluated using L'Hopital's rule or series expansion of trigonometric functions.

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