$\mathop {\lim }\limits_{h \to 0} \frac{{\sin \sqrt {x + h} - \sin \sqrt x }}{h}$ is equal to

$1/\left( {2\sin \sqrt x } \right)$

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}$, is

$\log \left( {\frac{b}{a}} \right)$

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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 691-720 of 930 questions

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691 Easy

691 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1+2+3+\ldots+n}{1+3+5+\ldots+(2 n-1)}=

View answer and explanation

Correct option: C

1 / 2

Option A (1/5) is correct because evaluating both limits separately and subtracting them gives 1/5.

692 Easy

692 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

\left( {\cos \sqrt x } \right)/2\sqrt x

Option C (

\left( {\cos \sqrt x } \right)/2\sqrt x

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

693 Medium

693 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{n+n}\right)=

View answer and explanation

Correct option: A

\log _e^2

Option A (

\log_e 2

) is correct because the given limit can be expressed as a definite integral, which evaluates to

\log_e 2

694 Medium

694 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\cos \left(\sin ^2 x\right)-\cos \left(x^2\right)}{x^6}

View answer and explanation

Correct option: C

1 / 3

Option C (

\frac{1}{3}

) is correct because using L'Hopital's rule or Taylor series expansion, the limit evaluates to 1/3.

695 Easy

695 of 930

Application +4 | -1

\lim\limits_{x \to \infty } \frac{{{{\left( {2x + 1} \right)}^{40}}{{\left( {4x - 1} \right)}^5}}}{{{{\left( {2x + 3} \right)}^{45}}}}

is equal to

View answer and explanation

Correct option: C

Option C (32) is correct because after factoring out

x

and simplifying, the limit evaluates to 32.

696 Medium

696 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: D

\sin 1

The correct answer is D (

\sin 1

) because as n approaches infinity,

\cos^n(1)

approaches 0 faster than

\sin^n(1)

, making the limit approximately

\sin(1)

697 Medium

697 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}(\cos x)^{\cot ^2 x}

View answer and explanation

Correct option: D

\frac{1}{\sqrt{e}}

Option D (

\frac{{1}}{{\sqrt{{e}}}}

) is correct because by applying L'Hôpital's rule to the logarithmic form of the limit, we find the limit to be

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

698 Easy

698 of 930

Analyzing +4 | -1

\lim \limits_{{h \rightarrow 0}}\left\{\frac{1}{h \cdot \sqrt[3]{8+h}}-\frac{1}{2 h}\right\}

is equal to

View answer and explanation

Correct option: D

-\frac{1}{48}

Option D (-1/48) is correct because after simplifying the expression and using L'Hopital's rule or series expansion, the limit evaluates to -1/48.

699 Medium

699 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow+\infty}}\left(\frac{2^x}{x^{10}}\right)

is equal to

View answer and explanation

Correct option: D

+\infty

Option D (

+\infty

) is correct because exponential functions grow faster than polynomial functions as x approaches infinity. Applying L'Hôpital's rule repeatedly confirms this.

700 Easy

700 of 930

Analyzing +4 | -1

The value of

\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}

View answer and explanation

Correct option: C

5 \sqrt{2}

The correct answer is C (

5 \sqrt{2}

) because by using L'Hopital's rule and trigonometric identities, the limit can be evaluated to

5\sqrt{2}

701 Easy

701 of 930

Application +4 | -1

\lim _{x \rightarrow 0} \frac{\sin 2 x}{x \cos x}

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because simplifying the expression using the double angle formula for sine and the limit of sin(x)/x as x approaches 0 leads to the answer 2.

702 Easy

702 of 930

Analyzing +4 | -1

The quadratic equation whose roots are the minimum value of

\sin ^2 \theta-\sin \theta+\frac{1}{2}

and

\lim \limits_{{x \rightarrow \infty}} \sqrt{(x+1)(x+2)}-x

View answer and explanation

Correct option: B

8 x^2-14 x+3=0

Option B,

8x^2 - 14x + 3 = 0

, is correct because the minimum value of the given trigonometric expression is

\frac{1}{4}

, and the limit of the given expression is

\frac{3}{2}

. Therefore, the quadratic equation with roots

\frac{1}{4}

and

\frac{3}{2}

8x^2 - 14x + 3 = 0

703 Hard

703 of 930

Analyzing +4 | -1

The set of all values of

a

for which

\lim \limits_{{x \rightarrow a}}([x-5]-[2 x+2])=0

, where

[\propto]

denotes the greatest integer less than or equal to

\propto

is equal to

View answer and explanation

Correct option: D

(-7.5,-6.5)

Option D ((

-7.5, -6.5

)) is correct because the set of values for which the limit equals zero is the open interval between -7.5 and -6.5.

704 Medium

704 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \left\{ {\frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}}} \right\}

View answer and explanation

Correct option: D

\frac{1}{{120}}

Option D (

\frac{1}{{120}}

) is correct because it is the result of evaluating the limit using the Taylor series expansion of sin(x).

705 Easy

705 of 930

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: A

\frac{3}{2}

Option A (

\frac{3}{2}

) is correct because the limit can be evaluated by using L'Hopital's rule or by using Taylor series expansions for

e^{x^2}

and

\cos x

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706 Easy

706 of 930

Application +4 | -1

The value of

\lim\limits_{x \to b} \frac{{\sqrt {x - a} - \sqrt {b - a} }}{{{x^2} - {b^2}}},

for

b>a

, is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because rationalizing the numerator and simplifying the expression leads to the given result.

707 Medium

707 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: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct. By dividing numerator and denominator by

x^{1/2}

and taking the limit as x approaches infinity, the expression simplifies to

\sqrt{2}

708 Easy

708 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}

, is

View answer and explanation

Correct option: A

\log \left( {\frac{a}{b}} \right)

Option A (

\log \left( {\frac{a}{b}} \right)

) is correct because it's a direct application of the limit definition of the derivative of an exponential function.

709 Easy

709 of 930

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} + 6}}{{{x^2} - 6}}} \right)^x}

is given by

View answer and explanation

Correct option: B

Option B (1) is correct because the limit can be evaluated by rewriting the expression and using the standard limit definition of

e

710 Easy

710 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{{\sin }^2}2x}}{{{{\sin }^2}4x}}

View answer and explanation

Correct option: C

\frac{1}{4}

Option C (1/4) is correct because by using the property

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

, the limit evaluates to 1/4.

711 Easy

711 of 930

Application +4 | -1

\lim _{x \rightarrow 0}\left(\frac{\log (6+x)-\log (6-x)}{x}\right)=

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

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

\lim_{x \to 0} \frac{\log(1+x)}{x} = 1

, the limit of the expression as x approaches 0 is 1/3.

712 Easy

712 of 930

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 can be evaluated by using the trigonometric identity

\sin(\pi - x) = \sin x

and the standard limit

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

713 Easy

713 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4

, then

View answer and explanation

Correct option: B

a=1, b=-4

Option B (

a=1, b=-4

) is correct because for the limit to be 4, the coefficients of

x^2

must be 0 and the ratio of the coefficients of

x

must be 4.

714 Easy

714 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because the limit simplifies to

\pi

after applying trigonometric identities and limit properties.

715 Easy

715 of 930

Application +4 | -1

Let

f:R \to R

be a differentiable function such that

f\left( 3 \right) = 3,f'\left( 3 \right) = \frac{1}{2}

, Then, the value of

\lim\limits_{x \to 3} \,\frac{{\int \nolimits_3^{f\left( x \right)} 2{t^3}\,dt}}{{x - 3}}

View answer and explanation

Correct option: C

Option C (27) is correct because applying L'Hopital's rule and the given values of f(3) and f'(3) leads to the result 27.

716 Easy

716 of 930

Analyzing +4 | -1

The value

\lim\limits_{x \to a} {\left( {\frac{{\sin x}}{{\sin a}}} \right)^{\frac{1}{{x - a}}}}

, is

View answer and explanation

Correct option: C

{e^{\cot a}}

Option C (

{e^{\cot a}}

) is correct because it follows from applying L'Hopital's rule after rewriting the limit in exponential form.

717 Easy

717 of 930

Understanding +4 | -1

Let

f(x)=\operatorname{sgn}(\operatorname{sgn}(\operatorname{sgn} x))

. Then

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand limit and right-hand limit are not equal, so the limit does not exist.

718 Easy

718 of 930

Understanding +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{a^n+b^n}{a^n-b^n}

, where

a > b > 1

, is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because dividing both numerator and denominator by

a^n

and using the fact that

(\frac{{b}}{{a}})^n

approaches 0 as n approaches infinity (since a > b > 1), the limit simplifies to 1.

719 Easy

719 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \,\,\frac{{{{\left( {\int \nolimits_0^x \,\,{e^x}\;\,dx} \right)}^2}}}{{\int \nolimits_0^x \,\,{e^{2{x^3}}}\;\,dx}}

, is

View answer and explanation

Correct option: D

\mathrm{0}

Option D (0) is correct because applying L'Hopital's rule repeatedly, the limit approaches 0 as the exponential term in the denominator grows much faster than the numerator.

720 Medium

720 of 930

Analyzing +4 | -1

The value of

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

View answer and explanation

Correct option: C

1/2

Option C (

1/2

) is correct because using the arctangent subtraction formula and then the small angle approximation for tangent yields the correct limit.

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