$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2\left( {x - 1} \right)} }}{{x - 1}}$

Exists and is equals $\sqrt 2 $

Exists and is equals $ - \sqrt 2 $

Does not exist because $x - 1 \to 0$

If $f(x)=\left\{\begin{array}{l}\frac{5}{2}-x, \text { when } x 2\end{array}\right.$, then:

$\lim _{x \rightarrow 2} f(x)=1$

Back to LIMITS

Your Progress

0 / 930 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →

Showing 811-840 of 930 questions

Preparing practice and saved progress…

811 Medium

811 of 930

Analyzing +4 | -1

a, b

and

c

are real numbers, then the value of

\lim \limits_{{t \rightarrow 0}} \ln \left(\frac{1}{t} \int_0^t(1+a \sin b x)^{c / x} d x\right)

equals

View answer and explanation

Correct option: A

a b c

Option A (

a b c

) is correct because the limit is evaluated using L'Hopital's rule and the exponential limit definition.

812 Easy

812 of 930

Analyzing +4 | -1

\lim\limits_{x \to \infty } \left\{ {\sqrt {{x^2} - x + 1} - ax - b} \right\} = 0

, then

View answer and explanation

Correct option: B

a = 1,\;b = - \frac{1}{2}

Option B (

a = 1, b = -\frac{1}{2}

) is correct because after rationalizing and comparing coefficients, these are the values that make the limit equal to 0.

813 Easy

813 of 930

Analyzing +4 | -1

\lim\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2\left( {x - 1} \right)} }}{{x - 1}}

View answer and explanation

Correct option: D

Does not exist because left hand limit is not equal to right hand limit

Option A (

{\left( {\log a} \right)^2}

) is correct. Applying L'Hopital's rule twice gives the result

{\left( {\log a} \right)^2}

814 Easy

814 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{1}{x} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)

View answer and explanation

Correct option: A

Option A (2) is correct because by using the trigonometric identity and limit properties, the given limit evaluates to 2.

815 Easy

815 of 930

Analyzing +4 | -1

n \in N

, and

[x]

denotes the greatest integer less than or equal to

x

, then

\lim\limits _{x \rightarrow n}(-1)^{[x]}

is equal to

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.

816 Medium

816 of 930

Analyzing +4 | -1

f

is a real-valued differentiable and "

a

" is a real constant. Further

f^{\prime}

is continuous. If

F(x)=\left|\begin{array}{ccc} f(x+a) & f(x+2 a) & f(x+3 a) \\ f(a) & f(2 a) & f(3 a) \\ f^{\prime}(a) & f^{\prime}(2 a) & f^{\prime}(3 a) \end{array}\right|

then

\lim \limits_{{x \rightarrow 0}} F(x) / x

View answer and explanation

Correct option: D

Option D (0) is correct because the limit evaluates to 0 after applying L'Hôpital's rule and using the properties of determinants.

817 Easy

817 of 930

Application +4 | -1

f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right)

is a continuous functions, then

f(\pi / 4)

is equal to -

View answer and explanation

Correct option: A

-\frac{1}{2}

Option A (-

\frac{1}{2}

) is correct because it is the limit of the function as x approaches

\frac{\pi}{4}

, ensuring continuity.

818 Easy

818 of 930

Application +4 | -1

Let

z_n=\frac{1^3+2^3+3^3+\cdots+n^3}{n^4}

Then

\lim _{n \rightarrow \infty} z_n

View answer and explanation

Correct option: C

\frac{1}{4}

Option C (

\frac{1}{4}

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

\frac{n^2(n+1)^2}{4}

, and dividing by

n^4

and taking the limit as

n

approaches infinity yields

\frac{1}{4}

819 Easy

819 of 930

Application +4 | -1

f(x)=\left\{\begin{array}{ccc}-2 \sin x & \text { if } & x \leq-\frac{\pi}{2} \\ A \sin x+B & \text { if } & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x & \text { if } & x \geq \frac{\pi}{2}\end{array}\right.

is a continuous function then

View answer and explanation

Correct option: B

A=-1, B=1

Option B (

A=-1, B=1

) is correct because these values ensure the function is continuous at

x=-\frac{\pi}{2}

and

x=\frac{\pi}{2}

820 Easy

820 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(1+\frac{a}{x}-\frac{4}{x^2}\right)^{2 x}=e^3

, then '

a

' is equal to

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{{3}}{{2}}

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

821 Medium

821 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow-\pi}} \frac{|x+\pi|}{\sin x}

is equal to

View answer and explanation

Correct option: D

Does not exist

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

822 Easy

822 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{1}{8 \sqrt{3}}

Option A (

\frac{1}{8 \sqrt{3}}

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

823 Easy

823 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}}(\cos a x)^{\operatorname{cosec}^2 b x}

View answer and explanation

Correct option: C

e^{\left(-\frac{a^2}{2 b^2}\right)}

Option C (

e^{\left(-\frac{a^2}{2 b^2}\right)}

) is correct because the limit can be solved by taking the natural logarithm of the expression and applying L'Hopital's rule or using trigonometric identities.

824 Easy

824 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}}\{\sqrt[3]{(x+a)(x+b)(x+c)}-x\}=

View answer and explanation

Correct option: B

\frac{a+b+c}{3}

Option B,

\frac{a+b+c}{3}

, is correct because it is the result of rationalizing the expression and then taking the limit.

825 Easy

825 of 930

Analyzing +4 | -1

f(x)=\left\{\begin{array}{r}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1, \text { when } x=a\end{array}\right.

, then:

View answer and explanation

Correct option: B

f(x)

is discontinuous at

x=a

Option B (

f(x)

is discontinuous at

x=a

) is correct because the left-hand limit and right-hand limit at

x=a

are not equal, indicating a discontinuity.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

826 Easy

826 of 930

Analyzing +4 | -1

f(x)=\left\{\begin{array}{l}\frac{5}{2}-x, \text { when } x<2 \\ 1, \text { when } x=2 \\ x-\frac{3}{2}, \text { when } x>2\end{array}\right.

, then:

View answer and explanation

Correct option: B

f(x)

is discontinuous at

x=2

The function f(x) is discontinuous at x=2 because the limit as x approaches 2 is 1/2, while f(2) = 1, thus they are not equal.

827 Medium

827 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}

View answer and explanation

Correct option: D

- 1/56

Option D (

\frac{-1}{56}

) is correct because it is the result of applying L'Hopital's rule to the limit after recognizing it as an indeterminate form.

828 Easy

828 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct. The limit is evaluated by converting the sum to an integral and then evaluating the integral.

829 Medium

829 of 930

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the limit evaluates to

\sqrt{\frac{5}{6}}

after simplification and using trigonometric limit properties.

830 Easy

830 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} x \log _e(\sin x)

is equal to

View answer and explanation

Correct option: B

\log _e 1

Option B (

\log_e 1

) is correct because the limit is evaluated using L'Hospital's Rule, resulting in 0, which is equal to

\log_e 1

831 Easy

831 of 930

Application +4 | -1

If the function

f(x)=\frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1}

is continuous at

x=0

then

f(0)

is equal to

View answer and explanation

Correct option: D

-4

Option D (-4) is correct because the limit of the function as x approaches 0 is -4, and for the function to be continuous at x=0, f(0) must equal this limit.

832 Easy

832 of 930

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left\{\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n+1)(2 n+3)}\right\},

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because the series telescopes, and the limit as

n

approaches infinity is

\frac{1}{2}

833 Easy

833 of 930

Analyzing +4 | -1

Let

\beta = \lim\limits_{x \to 0} \frac{{\alpha x - \left( {{e^{3x}} - 1} \right)}}{{\alpha x\left( {{e^{3x}} - 1} \right)}}

. Then the value of

\alpha+\beta

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because it involves using L'Hopital's rule or series expansion to evaluate the limit and then summing

\alpha

and

\beta

834 Easy

834 of 930

Analyzing +4 | -1

\lim\limits_{n \to \infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + \frac{3}{{1 - {n^2}}} + \ldots + \frac{n}{{1 - {n^2}}}} \right\}

is equal to

View answer and explanation

Correct option: B

\frac{{ - 1}}{2}

The correct answer is B (

\frac{{ - 1}}{2}

) because the limit is evaluated by summing the series and then simplifying the expression.

835 Medium

835 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} {\left( {\cos x} \right)^{{{\cot }^2}x}}

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the result of evaluating the limit using L'Hopital's rule or standard limit approximations.

836 Easy

836 of 930

Analyzing +4 | -1

Let

\alpha, \beta \in R

be such that

\lim \limits_{{x \rightarrow 0}} \frac{x^2 \sin \beta x}{\alpha x-\sin x}=1

. Then,

6(\alpha+\beta)

equals

View answer and explanation

Correct option: B

Option B (7) is correct because by applying L'Hopital's rule and standard limits, we find

\alpha = 1

and

\beta = \frac{{1}}{{6}}

, thus

6(\alpha + \beta) = 7

837 Easy

837 of 930

Analyzing +4 | -1

The integer '

n

' for which the

\lim \limits_{{x \rightarrow 0}} \frac{\cos ^2 x-\cos x-e^x \cos x+e^x-\frac{x^3}{2}}{x^n}

is a finite non-zero number, is

View answer and explanation

Correct option: C

Option C (4) is correct because when n=4, the limit evaluates to a finite non-zero number after expanding the trigonometric and exponential functions using their Maclaurin series.

838 Medium

838 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)^2}{2 x \tan x-x \tan 2 x}

is :

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because after using trigonometric expansions and simplifying, the limit evaluates to -2.

839 Easy

839 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{x\left(e^{\frac{\sqrt{1+x^2+x^4}-1}{x}}-1\right)}{\sqrt{1+x^2+x^4}-1}

View answer and explanation

Correct option: B

Is equal to 1

Option B (Is equal to 1) is correct because the limit is evaluated using substitution and standard limit results, resulting in 1.

840 Easy

840 of 930

Analyzing +4 | -1

\lim\limits_{h \to 0} \frac{{f\left( {2h + 2 + {h^2}} \right) - f\left( 2 \right)}}{{f\left( {h - {h^2} + 1} \right) - f\left( 1 \right)}}

, given that

f'\left( 2 \right) =

and

f'\left( 1 \right) = 4

View answer and explanation

Correct option: D

is equal to 3

Option D (is equal to 3) is correct because applying L'Hopital's rule and substituting the given derivatives yields the correct result.

Revision notes and key takeaways

Review LIMITS in your textbook or class notes, then use this set to check your understanding.

Showing 811-840 of 930 questions

Your Progress

Limits practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice