Given that $\mathop {\lim }\limits_{n \to \infty } \mathop \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}$, is

${\left( {\frac{e}{4}} \right)^k}$

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JEE • MATHEMATICS • LIMITS

Limits practice · Hard

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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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1 Hard

1 of 16

Analyzing +4 | -1

Let

f\left( x \right)

be a differentiable function such that

f\left( 1 \right) = 2

. If

\lim\limits_{x \to 1} \,\,\int \limits_2^{f\left( x \right)} \,\,\frac{{2t}}{{x - 1}}\,\,dt = 4

, then the value of

f'\left( 1 \right)

View answer and explanation

Correct option: A

\mathrm{1}

Option A (1) is correct because applying L'Hopital's rule and the Fundamental Theorem of Calculus leads to

f'(1) = 1

2 Hard

2 of 16

Application +4 | -1

Let

f:R \to R

be a differentiable function such that

f\left( 2 \right) = 2

. Then, the value of

\lim\limits_{x \to 2} \int \limits_2^{f\left( x \right)} \frac{{4\;{t^3}}}{{x - 2}}dt

, is

View answer and explanation

Correct option: C

32f'\left( 2 \right)

Option C (

32f'(2)

) is correct because it is the result of applying L'Hopital's rule and the fundamental theorem of calculus to the given limit.

3 Hard

3 of 16

Analyzing +4 | -1

f(x)=\frac{e^{[x]+\mid x \mid}-3}{[x]+|x|+1}

, then: (where [.] represents greatest integer function)

View answer and explanation

Correct option: A

\lim \limits_{x \rightarrow 0^{+}} f(x)=-2

Option A is correct because the right-hand limit exists and equals -2, while the left-hand limit does not exist.

4 Hard

4 of 16

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17

, then

5 a^{2}+b^{2}

is equal to

View answer and explanation

Correct option: C

Option C (68) is correct because it is obtained by using L'Hopital's rule and Taylor series expansion to evaluate the limit and then solving for the expression.

5 Hard

5 of 16

Analyzing +4 | -1

g\left( x \right)

is a polynomial satisfying

g\left( x \right)g\left( y \right) = g\left( x \right) + g\left( y \right) + g\left( {xy} \right) - 2

for all real

x

and

y

and

g\left( 2 \right) = 5

, then

\lim\limits_{x \to 3} g\left( x \right)

View answer and explanation

Correct option: B

Option B (10) is correct because by substituting the given condition into the functional equation and solving for g(x), we find that g(x) = x² + 1, and thus

\lim\limits_{x \to 3} g(x) = 10

6 Hard

6 of 16

Analyzing +4 | -1

For each

x \in R

, let

[x]

be the greatest integer less than or equal to

x

. Then,

\lim\limits_{x \to {0^ - }} \frac{{x([x] + \left| x \right|)\sin [x]}}{{\left| x \right|}}

is equal to

View answer and explanation

Correct option: C

- sin 1

Option C (- sin 1) is correct because it is the limit of the given expression as x approaches 0 from the left, obtained by substituting the greatest integer function and absolute value.

7 Hard

7 of 16

Application +4 | -1

The value of

\underset{n \rightarrow \infty}{\operatorname{Lim}}\left(\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots .+\frac{1}{n+n m}\right)

is equal to
[Note : where

\mathrm{m}>0

View answer and explanation

Correct option: C

\frac{\ln (1+m)}{m}

Option C (

\frac{\ln (1+m)}{m}

) is correct because it correctly evaluates the limit of the Riemann sum as a definite integral.

8 Hard

8 of 16

Application +4 | -1

\lim _{n \rightarrow \infty} \sum\limits_{{r=1}}^{n} \frac{1}{n} e^{\frac{r}{n}}

View answer and explanation

Correct option: B

e-1

Option B (e-1) is correct because the limit of the Riemann sum can be evaluated as a definite integral, which simplifies to e-1.

9 Hard

9 of 16

Analyzing +4 | -1

\lim\limits_{{x \rightarrow -\infty}}\frac{x^5 \tan \left(\frac{1}{\pi x^2}\right)+3|x|^2+7}{|x|^3+7|x|+8}

is equal to 

View answer and explanation

Correct option: A

-\frac{1}{\pi}

Option A (-1/π) is correct because after dividing by |x|^3 and using the limit tan(θ)/θ as θ approaches 0, the expression simplifies to -1/π.

10 Hard

10 of 16

Application +4 | -1

Given that

\lim\limits_{n \to \infty } \sum \limits_{r = 1}^n \frac{{\log \left( {r + n} \right) - \log n}}{n} = 2\left( {\log 2 - \frac{1}{2}} \right),\lim\limits_{n \to \infty } \frac{1}{{{n^k}}}{\left[ {{{\left( {n + 1} \right)}^k}{{\left( {n + 2} \right)}^k} \ldots {{\left( {n + n} \right)}^k}} \right]^{1/n}}

, is

View answer and explanation

Correct option: C

{\left( {\frac{4}{e}} \right)^k}

Option C (

{\left( {\frac{4}{e}} \right)^k}

) is correct because the limit can be expressed as an integral and evaluated using the given information.

11 Hard

11 of 16

Analyzing +4 | -1

\lim \limits_{{x \rightarrow-1}} \frac{\cos 2-\cos 2 x}{x^2-|x|}

is equal to

View answer and explanation

Correct option: C

2 \sin 2

Option C (

2 \sin 2

) is correct because the limit is evaluated using trigonometric identities and limit properties, resulting in

2 \sin 2

12 Hard

12 of 16

Analyzing +4 | -1

\lim\limits_{x \to \frac{\pi }{2}} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}},a > 0

is equal to

View answer and explanation

Correct option: C

{\log _e}a

Option C (

{\log _e}a

) is correct because, by using L'Hopital's rule or series expansion, the limit evaluates to the natural logarithm of a.

13 Hard

13 of 16

Analyzing +4 | -1

\lim\limits_{x \to 3} \,\left( {\left[ {x - 3} \right] + \left[ {3 - x} \right] - x} \right),

where

\left[ \cdot \right]

denotes the greatest integer function, is equal to

View answer and explanation

Correct option: B

-4

Option B (-4) is correct because the greatest integer function simplifies to -1, and the limit can be directly evaluated.

14 Hard

14 of 16

Analyzing +4 | -1

\lim\limits_{x \to 1} \frac{{{x^2} - ax + b}}{{x - 1}} = 5

, then a+b is equal to

View answer and explanation

Correct option: C

-7

Option C (-7) is correct because by applying L'Hôpital's rule or factoring and simplifying, we find that a+b = -7.

15 Hard

15 of 16

Analyzing +4 | -1

The set of all values of

a

for which

\lim \limits_{{x \rightarrow a}}([x-5]-[2 x+2])=0

, where

[\propto]

denotes the greatest integer less than or equal to

\propto

is equal to

View answer and explanation

Correct option: D

(-7.5,-6.5)

Option D ((

-7.5, -6.5

)) is correct because the set of values for which the limit equals zero is the open interval between -7.5 and -6.5.

Explore Other Chapters in MATHEMATICS

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

16 Hard

16 of 16

Application +4 | -1

\mathop {{\text{Lim}}}\limits_{x \to 0} \left( {1 + 2 + 3 + ........ + \left[ {\frac{1}{{\left| {\text{x}} \right|}}} \right]} \right)

where [.] denotes greatest integer function , is equal to

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct. By evaluating the limit as x approaches 0, the expression converges to

\frac{1}{2}

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 16 single-correct-answer MCQs on LIMITS.
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