Let $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^2}}&{{\text{if}}}&{x 3} \end{array}} \right.$ Which one of the following is not correct?

$\lim _{x \rightarrow 2} f(x)=4$

$\lim _{x \rightarrow 3} f(x)=2$

$\lim _{x \rightarrow 4} f(x)=4$

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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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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 647 questions

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601 Easy

601 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 2}} \frac{\sum\limits_{{r=1}}^n x^r-\sum\limits_{{r=1}}^n 2^r}{x-2}

is equal to

View answer and explanation

Correct option: C

(n-1) 2^n+1

Option C,

(n-1) 2^n+1

, is correct because it is the result of differentiating the sums, applying the limit, and then simplifying the resulting series.

602 Easy

602 of 647

Understanding +4 | -1

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

, us

View answer and explanation

Correct option: B

Option B (0) is correct because x² approaches 0, and sin(π/x) is bounded between -1 and 1, so their product approaches 0.

603 Easy

603 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow-1}} \frac{x^2+3 x+2}{x^2+4 x+3}

View answer and explanation

Correct option: D

1 / 2

Option D (

1 / 2

) is correct because the limit is found by factoring and simplifying the rational expression before evaluating at x = -1.

604 Easy

604 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(1+\frac{a \sin b x}{\cos x}\right)^{\frac{1}{x}}

, where

a, b

are non-zero constants is equal to:

View answer and explanation

Correct option: C

e^{a b}

Option C (e^(ab)) is correct because the limit is of the form 1^infinity, which simplifies to e^(ab) using the standard limit formula.

605 Easy

605 of 647

Application +4 | -1

\lim\limits_{x \to \pi /4} \frac{{1 - {{\cot }^3}x}}{{2 - \cot x - {{\cot }^3}x}}

, is

View answer and explanation

Correct option: B

\frac{3}{4}

Option B (

\frac{3}{4}

) is correct because it is the result of factoring and simplifying the expression after substituting

y = \cot x

and then evaluating the limit.

606 Easy

606 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{3x + \left| x \right|}}{{7x - \left| x \right|}}

, is

View answer and explanation

Correct option: D

None of these

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

607 Easy

607 of 647

Analyzing +4 | -1

\lim \limits_{{y \rightarrow 0}} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}

View answer and explanation

Correct option: A

Exists and equals

\frac{1}{4 \sqrt{2}}

Option A (Exists and equals

\frac{1}{4 \sqrt{2}}

) is correct because after rationalizing the numerator twice, the limit can be evaluated to

\frac{1}{4 \sqrt{2}}

608 Easy

608 of 647

Application +4 | -1

Let

f: {R} \rightarrow {R}

be a continuous function such that

f(3 x)-f(x)=x

. If

f(8)=7

, then

f(14)

is equal to

View answer and explanation

Correct option: B

Option B (10) is correct because by using the given functional equation and the limit concept, we can find f(0) and then f(14).

609 Easy

609 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{2 a x+(a-1) \sin x}{\tan ^3 x}=l

, then

a+l

is equal to

View answer and explanation

Correct option: D

\frac{4}{9}

Option D (4/9) is correct because by applying L'Hopital's rule and standard limits, we find

a = \frac{{1}}{{3}}

and

l = \frac{{1}}{{9}}

, thus

a + l = \frac{{4}}{{9}}

610 Easy

610 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 2}} \frac{\sqrt{x+7}-3 \sqrt{2 x-3}}{\sqrt[3]{x+6}-2 \sqrt[3]{3 x-5}}

View answer and explanation

Correct option: B

\frac{34}{23}

Option B (

\frac{34}{23}

) is correct because the limit is evaluated using L'Hôpital's Rule or by rationalizing and simplifying the expression.

611 Easy

611 of 647

Application +4 | -1

\lim _{n \rightarrow \infty}\left\{\frac{1+\frac{1}{3}+\frac{1}{9}+\cdots \text { to }(n+1) \text { terms }}{1+\frac{1}{2}+\frac{1}{4}+\cdots \text { to }(n+1) \text { terms }}\right\}=

View answer and explanation

Correct option: D

\frac{2}{3}

Option A (2/3) is correct because the limit simplifies to 2/3 after evaluating the geometric series and taking the limit as n approaches infinity.

612 Easy

612 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 2}} \frac{e^{3 x-6}-1}{\sin (2-x)}

, is

View answer and explanation

Correct option: C

-3

Option C (-3) is correct because the limit can be evaluated using the standard limit

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

and

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

613 Easy

613 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\log _e(1+x)-x}{x\left\{(1+x)^{1 / x}-e\right\}}

equals to

View answer and explanation

Correct option: C

1 / e

Option C (

\frac{1}{e}

) is correct because by using Taylor series expansion for

\log_e(1+x)

and

(1+x)^{1/x}

, we can simplify the limit and find its value.

614 Easy

614 of 647

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because the limit can be rewritten in the form of

e^{\lim_{x \to \infty} f(x)}

and then evaluated.

615 Easy

615 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow \pi / 2}}(\sec x-\tan x)

is equal to

View answer and explanation

Correct option: A

Option A is correct because the limit of (sec x - tan x) as x approaches π/2 is 0.

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616 Easy

616 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{\frac{1}{2}(1-\cos 2 x)}}{x}

, is

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.

617 Easy

617 of 647

Analyzing +4 | -1

Let

f(x)=x(-1)^{[1 / x]}, x \neq 0

, where

[x]

denotes the greatest integer less than or equal to

x

, then

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

View answer and explanation

Correct option: C

Is equal to 0

Option C (Is equal to 0) is correct because as x approaches 0, the function is bounded by x, making the limit 0.

618 Easy

618 of 647

Analyzing +4 | -1

f\left( x \right)

is differentiable and strictly increasing function, then the value of

\lim\limits_{x \to 0} \frac{{f\left( {{x^2}} \right) - f\left( x \right)}}{{f\left( x \right) - f\left( 0 \right)}}

View answer and explanation

Correct option: C

- 1

Option C (-1) is correct because applying L'Hopital's rule and evaluating the limit results in -1.

619 Easy

619 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left(\sqrt{n^2+n+1}-\left[\sqrt{n^2+n+1}\right]\right)

, where [.] denotes the greatest integer function is

View answer and explanation

Correct option: B

1 / 2

Option B (1/2) is correct because the expression simplifies to

\frac{1}{2}

after rationalizing and taking the limit as n approaches infinity.

620 Easy

620 of 647

Analyzing +4 | -1

{S_n} = \sum \limits_{k = 1}^n \,\,{a_k}

and

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

, then

\lim\limits_{n \to \infty } \frac{{{S_{n + 1}} - {S_n}}}{{\sqrt {\sum \nolimits_{k = 1}^n \,k} }}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the limit of the given expression as n approaches infinity is 0.

621 Easy

621 of 647

Application +4 | -1

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

, where [.] denotes the greatest integer function, is

View answer and explanation

Correct option: B

\frac{8}{3}

Option B (

5 \sqrt{3}

) is correct because the area of a parallelogram is half the magnitude of the cross product of its diagonals.

622 Easy

622 of 647

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2}}&{{\text{if}}}&{x < 1} \\ {x + 2}&{{\text{if}}}&{1 \leqslant x \leqslant 2} \\ {8 - 2x}&{{\text{if}}}&{2 < x \leqslant 3} \\ {2x - 4}&{i{\text{f}}}&{x > 3} \end{array}} \right.

Which one of the following is not correct?

View answer and explanation

Correct option: A

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

Option A is correct because the left-hand limit at x=1 is 1, but the right-hand limit is 3, therefore the limit does not exist, making the statement incorrect.

623 Easy

623 of 647

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.

624 Easy

624 of 647

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.

625 Easy

625 of 647

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.

626 Easy

626 of 647

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.

627 Easy

627 of 647

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.

628 Easy

628 of 647

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.

629 Easy

629 of 647

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).

630 Easy

630 of 647

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.

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