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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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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 631-647 of 647 questions

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631 Easy

631 of 647

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.

632 Easy

632 of 647

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.

633 Easy

633 of 647

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.

634 Easy

634 of 647

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.

635 Easy

635 of 647

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

636 Easy

636 of 647

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.

637 Easy

637 of 647

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}

638 Easy

638 of 647

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

639 Easy

639 of 647

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.

640 Easy

640 of 647

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.

641 Easy

641 of 647

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.

642 Easy

642 of 647

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.

643 Easy

643 of 647

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}}}

644 Easy

644 of 647

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}}

645 Easy

645 of 647

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.

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646 Easy

646 of 647

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.

647 Easy

647 of 647

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)

Revision notes and key takeaways

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