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 · 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 241-270 of 930 questions

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241 Easy

241 of 930

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.

242 Medium

242 of 930

Application +4 | -1

The derivative of

\cos \left(x^2+1\right)

by using first principle is

View answer and explanation

Correct option: B

-2 x \sin \left(x^2+1\right)

Option B (

-2 x \sin \left(x^2+1\right)

) is correct because it is the derivative of

\cos \left(x^2+1\right)

obtained using the first principle of differentiation and trigonometric identities.

243 Easy

243 of 930

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.

244 Easy

244 of 930

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.

245 Medium

245 of 930

Analyzing +4 | -1

Let

f(x)=\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^2}

for

x \neq \pi / 6

. Then

\lim _{x \rightarrow \pi / 6} f(x)

View answer and explanation

Correct option: B

\frac{1}{36}

Option B (

\frac{1}{36}

) is correct. By using trigonometric identities and L'Hopital's rule, the limit can be evaluated to

\frac{1}{36}

246 Easy

246 of 930

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.

247 Medium

247 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} x\log \sin x

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because by rewriting the expression and applying L'Hopital's rule twice, the limit evaluates to 0.

248 Easy

248 of 930

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.

249 Easy

249 of 930

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.

250 Easy

250 of 930

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.

251 Medium

251 of 930

Analyzing +4 | -1

If [x] denotes the greatest integer less than or equal to

x

, then the value of

\lim\limits_{x \to 1} \left\{ {1 - x + \left[ {x - 1} \right] + \left[ {1 - x} \right]} \right\}

View answer and explanation

Correct option: C

- 1

Option C (-1) is correct because both the left-hand limit and the right-hand limit as x approaches 1 are equal to -1, considering the greatest integer function.

252 Easy

252 of 930

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.

253 Easy

253 of 930

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.

254 Easy

254 of 930

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.

255 Medium

255 of 930

Application +4 | -1

The value of

\lim \limits_ {x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}

View answer and explanation

Correct option: A

Option A (1) is correct because the limit can be evaluated by using L'Hôpital's Rule after transforming the expression into an exponential form.

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256 Medium

256 of 930

Application +4 | -1

Let

f(x)= \begin{cases}1 /|x| & \text { for }|x| \geq 1 \\ a x^2+b & \text { for }|x|<1\end{cases}

The coefficients

a

and

b

so that

f

is continuous and differentiable at any point, are equal to

View answer and explanation

Correct option: A

a=-1 / 2, b=3 / 2

Option A (

a=-1 / 2, b=3 / 2

) is correct because these values ensure that the function is both continuous and differentiable at

x = 1

and

x = -1

257 Easy

257 of 930

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.

258 Medium

258 of 930

Application +4 | -1

Let

f(x)

be a differentiable function at

x=a,

f^1(a)=4,

f(a)=2.

Then

\lim \limits_{{x \rightarrow a}} \frac{x f(a)-a f(x)}{x-a}

is equal to

View answer and explanation

Correct option: B

2-4 a

Option B (

2-4a

) is correct because applying L'Hopital's rule and using the given values of

f(a)

and

f'(a)

leads to this result.

259 Easy

259 of 930

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.

260 Easy

260 of 930

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.

261 Easy

261 of 930

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.

262 Easy

262 of 930

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}}

263 Easy

263 of 930

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.

264 Easy

264 of 930

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.

265 Medium

265 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{4+\sin 2 x+A \sin x+B \cos x}{x^2}

exists, then the values of

A

and

B

are

View answer and explanation

Correct option: A

-2

and

-4

The correct answer is A (-2 and -4) because for the limit to exist, both the numerator and its derivative must approach zero as x approaches zero, leading to A = -2 and B = -4.

266 Easy

266 of 930

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.

267 Easy

267 of 930

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.

268 Easy

268 of 930

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.

269 Medium

269 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\log (\sin 7 x+\cos 7 x)}{\sin 3 x}

equals :

View answer and explanation

Correct option: B

\frac{7}{3}

Option B (

\frac{7}{3}

) is correct. By using L'Hopital's rule or series expansion, the limit evaluates to

\frac{7}{3}

270 Medium

270 of 930

Application +4 | -1

\Delta(x)=\left|\begin{array}{cc}e^x & \sin x \\ \cos x & \ln \left(1+x^2\right)\end{array}\right|

, then the value of

\lim _{x \rightarrow 0} \frac{\Delta(x)}{x}

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because by applying L'Hôpital's rule or using standard limits, the limit evaluates to -1.

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