If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}$, where $r = 1,2,3, \ldots ,n$, then $\mathop {\lim }\limits_{n \to \infty } {z_1}{z_2} \ldots {z_n}$ is equal to

$\cos \left( {\frac{\alpha }{2}} \right) - i\sin \left( {\frac{\alpha }{2}} \right)$

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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 151-180 of 647 questions

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151 Easy

151 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{a x-\left(e^{4 x}-1\right)}{a x\left(e^{4 x}-1\right)}=b

exists finitely then

2(a+b)=

View answer and explanation

Correct option: D

The correct answer is D (7) because by applying L'Hopital's rule and the condition for the limit to exist finitely, we find a = 4 and b = -1/2, thus 2(a+b) = 7.

152 Easy

152 of 647

Analyzing +4 | -1

Let

k > 0

and

\lambda=\lim \limits_{{x \rightarrow 0}} \frac{k\left(1-4 \sqrt{k^2-x^2}\right)}{x^2 \sqrt{k^2-x^2}}

be finite. Then the value of

\lambda k

, is

View answer and explanation

Correct option: B

\lambda=8, k=\frac{1}{4}

Option B (

\lambda=8, k=\frac{1}{4}

) is correct because the limit exists only when k = 1/4, and substituting this value gives lambda = 8.

153 Easy

153 of 647

Application +4 | -1

The value of

\lim\limits_{n \to \infty } \cos \left( {\frac{x}{2}} \right)\cos \left( {\frac{x}{4}} \right)\cos \left( {\frac{x}{8}} \right) \ldots \cos \left( {\frac{x}{{{2^n}}}} \right)

View answer and explanation

Correct option: C

\frac{{\left( {\sin x} \right)}}{x}

Option C (

\frac{{\left( {\sin x} \right)}}{x}

) is correct because using the product of cosines identity and taking the limit as n approaches infinity results in this expression.

154 Easy

154 of 647

Analyzing +4 | -1

The value of

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

, 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.

155 Easy

155 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sin 2 x}{x}

is equal to

View answer and explanation

Correct option: A

Option A is correct because the limit evaluates to 2.

156 Easy

156 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \left\{ {\frac{1}{{x\sqrt[3]{{8 + x}}}} - \frac{1}{{2x}}} \right\}

is equal to

View answer and explanation

Correct option: D

\frac{{ - 1}}{{48}}

Option D (

\frac{{ - 1}}{{48}}

) is correct because it's the result of applying L'Hopital's rule or series expansion to evaluate the limit.

157 Easy

157 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{1 - \cos \left( {1 - \cos x} \right))}}{{{x^4}}}

, is

View answer and explanation

Correct option: A

\frac{1}{8}

Option A (1/8) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 1/8.

158 Easy

158 of 647

Application +4 | -1

The value of

\lim\limits_{x \to \pi /2} {\left( {\sin x} \right)^{\tan x}}

, is

View answer and explanation

Correct option: B

Option B (1) is correct because by applying L'Hopital's rule after taking the natural logarithm, the limit can be evaluated to 0, and e⁰ = 1.

159 Easy

159 of 647

Application +4 | -1

\lim _{x \rightarrow 0} \int_0^x \frac{t \sin (10 t) d t}{x}

is equal to:

View answer and explanation

Correct option: A

Option A (0) is correct because applying L'Hôpital's rule results in a limit of 0.

160 Easy

160 of 647

Application +4 | -1

\lim\limits_{n \to \infty } \sum\nolimits_{r = 1}^n {\frac{1}{n}{e^{r/n}}}

View answer and explanation

Correct option: B

e - 1

Option B (e - 1) is correct because the limit of the Riemann sum is equal to the definite integral of e^x from 0 to 1, which evaluates to e - 1.

161 Easy

161 of 647

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: D

\frac{5}{2}

Option D (

\frac{5}{2}

) is correct because the limit can be evaluated using L'Hopital's rule or by expanding the exponential and cosine functions into their Taylor series.

162 Easy

162 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}

is equal to

View answer and explanation

Correct option: A

-1

Option A (-1) is correct because the limit is evaluated using standard limit results, resulting in -1.

163 Easy

163 of 647

Application +4 | -1

\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{{1}}{{4}}

) is correct because after simplifying the expression inside the limit, we can directly substitute x = 2 to find the limit.

164 Easy

164 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{\log x}{[x]}

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

View answer and explanation

Correct option: A

Option A (0) is correct because the limit is evaluated using the squeeze theorem and L'Hopital's rule.

165 Easy

165 of 647

Application +4 | -1

{z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}

, where

r = 1,2,3, \ldots ,n

, then

\lim\limits_{n \to \infty } {z_1}{z_2} \ldots {z_n}

is equal to

View answer and explanation

Correct option: C

{e^{i\alpha /2}}

Option C (

{e^{i\alpha /2}}

) is correct because by using Euler's formula and simplifying the sum, the limit converges to

{e^{i\alpha /2}}

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166 Easy

166 of 647

Analyzing +4 | -1

f(x)=\frac{1}{3}\left\{f(x+1)+\frac{5}{f(x+2)}\right\}

and

f(x) > 0

for all

x \in R

, then

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

View answer and explanation

Correct option: B

\sqrt{\frac{5}{2}}

Option B (

\sqrt{\frac{5}{2}}

) is correct because the limit can be found by assuming the limit exists and solving for it.

167 Easy

167 of 647

Application +4 | -1

f\left( a \right) = 2,f'\left( a \right) = 1,g\left( a \right) = - 1,g'\left( a \right) = 2

, then the value of

\lim\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}

, is

View answer and explanation

Correct option: C

Option A (-28) is correct because by applying L'Hopital's rule, the limit can be evaluated to -28.

168 Easy

168 of 647

Analyzing +4 | -1

\lim\limits_{x \to - 3} \frac{{3{x^2} + ax + a - 7}}{{{x^2} + 2x - 3}}

exists, then

a

is equal to

View answer and explanation

Correct option: A

The correct answer is A (10) because for the limit to exist, the numerator must also approach zero as x approaches -3, allowing for factorization and simplification.

169 Easy

169 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \pi / 2}} \frac{\left[\frac{x}{2}\right]}{\log (\sin x)}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because as x approaches

\pi/2

, the greatest integer function of x/2, denoted as [x/2], becomes 0, making the entire limit equal to 0.

170 Easy

170 of 647

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the product of a function approaching zero and a bounded function is zero.

171 Easy

171 of 647

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

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

172 Easy

172 of 647

Analyzing +4 | -1

\lim\limits_{n \to \infty } \frac{{{2^{ - n}}\left( {{n^2} + 5n + 6} \right)}}{{\left( {n + 4} \right)\left( {n + 5} \right)}}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because as n approaches infinity, the exponential term 2^(-n) dominates, causing the entire expression to approach 0.

173 Easy

173 of 647

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because the sine function is bounded between -1 and 1, and the

x^2

term approaches 0 as x approaches 0, making the entire expression approach 0.

174 Easy

174 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sin (\sin x)-\sin x}{a x^3+b x^5+c}=-\frac{1}{12}

, then

View answer and explanation

Correct option: A

a=2, b \in R, c=0

Option A (

a=2, b \in R, c=0

) is correct because the limit exists and equals -1/12 only when a=2 and c=0, with b being any real number.

175 Easy

175 of 647

Application +4 | -1

The value of

\lim\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}}

, is

View answer and explanation

Correct option: A

{e^2}

Option A (

{e^2}

) is correct because it is the result of evaluating the limit using the exponential limit formula.

176 Easy

176 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2+4 x-3}{x^2-2 x+5}\right)^x

is equal to

View answer and explanation

Correct option: A

e^6

Option A (

e^6

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

\lim_{x \rightarrow \infty} (1 + \frac{{a}}{{x}})^x = e^a

177 Easy

177 of 647

Analyzing +4 | -1

\lim\limits_{x \to \infty } \frac{{\sqrt {{x^2} + 1} - \sqrt[3]{{{x^3} + 1}}}}{{\sqrt[4]{{{x^4} + 1}} - \sqrt[5]{{{x^4} + 1}}}}

equals

View answer and explanation

Correct option: B

Option B (0) is correct because by dividing the numerator and denominator by the highest power of x and simplifying, the limit evaluates to 0.

178 Easy

178 of 647

Application +4 | -1

f(1)=1

and

f^{\prime}(1)=4

, then the value of

\lim _{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1}

View answer and explanation

Correct option: B

Option B (4) is correct because by applying L'Hôpital's rule, the limit simplifies to

\frac{{f'(1)}}{{1 / (2 \sqrt{{1}})}} = \frac{{4}}{{1/2}} = 4

179 Easy

179 of 647

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because after applying L'Hopital's rule or using series expansion, the limit evaluates to 1.

180 Easy

180 of 647

Analyzing +4 | -1

\left|\cos ^{-1} \frac{1}{n}\right| < \frac{\pi}{2}

, then

\lim \limits_{{n \rightarrow \infty}}\left\{(n+1) \frac{2}{\pi} \cos ^{-1} \frac{1}{n}-n\right\}

is equal to

View answer and explanation

Correct option: B

\frac{\pi-2}{\pi}

Option B (

\frac{\pi-2}{\pi}

) is correct because after applying trigonometric identities and evaluating the limit, the expression simplifies to

\frac{\pi-2}{\pi}

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