The value of $\mathop {\lim }\limits_{x \to \infty } {\left\{ {\frac{{a_1^{1/x} + a_2^{1/x} + \ldots + a_n^{1/x}}}{n}} \right\}^1}$, is

${a_1} + {a_2} + \ldots + {a_n}$

${e^{{a_1} + {a_2} + \ldots + {a_n}}}$

$\frac{{{a_1} + {a_2} + \ldots + {a_n}}}{n}$

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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 61-90 of 930 questions

61 Easy

61 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to \pi /2} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}

View answer and explanation

Correct option: B

\frac{1}{{16}}

Option B (

\frac{1}{16}

) is correct because by substituting

x = \frac{\pi}{2} + h

and simplifying the trigonometric functions, the limit can be evaluated to

\frac{1}{16}

62 Easy

62 of 930

Application +4 | -1

Given

f\left( x \right) = \frac{{ax + b}}{{x + 1}},\lim\limits_{x \to \infty } f\left( x \right) = 1

and

\lim\limits_{x \to 0} f\left( x \right) = 2

, then

f\left( { - 2} \right)

View answer and explanation

Correct option: A

Option A (0) is correct because by using the given limits, we find a=1, b=2, and substituting these values into f(-2) results in 0.

63 Easy

63 of 930

Analyzing +4 | -1

Let

f(x) = \left\{ {\begin{array}{llllllllllllllllllll} {x - 1}&{{\text{ if }}x \leqslant 1} \\ {2x - 1}&{{\text{ if }}1 < x < 2} \\ {x + 1}&{{\text{ if }}x \geqslant 2} \end{array}} \right.\,

then

\lim\limits_{x \to 1} f\left( x \right)

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 at x=1 are not equal, indicating the limit does not exist.

64 Medium

64 of 930

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}}}

View answer and explanation

Correct option: A

{e^2}

Option A (

{e^2}

) is correct because the limit can be transformed into the form

{e^{\lim f(x)}}

and the exponent evaluates to 2.

65 Easy

65 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because by substituting

x = \tan \theta

and using trigonometric identities, the limit simplifies to 2.

66 Easy

66 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\int\limits_{0}^{x^2} \sec ^2 t d t}{x \sin x} d x

, is

View answer and explanation

Correct option: D

Option D, 1, is correct because it is the result of applying L'Hopital's rule to the limit and simplifying the expression.

67 Easy

67 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}}\left(1+\tan ^2 \sqrt{x}\right)^{1 / 2 x}

View answer and explanation

Correct option: B

e^{1 / 2}

Option B (

e^{1 / 2}

) is correct because

\lim_{x \rightarrow 0} (1+\tan^2 \sqrt{x})^{1 / 2x} = e^{\lim_{x\rightarrow 0} (\tan^2 \sqrt{x})/(2x)} = e^{1/2}

68 Easy

68 of 930

Analyzing +4 | -1

f(x)=\left\{\begin{array}{r}x^2, \text { when } x \neq 1 \\ 2, \text { when } x=1\end{array}\right.

, then:

View answer and explanation

Correct option: C

f(x)

is discontinuous at

x=1

Option C (

f(x)

is discontinuous at

x=1

) is correct because the limit as x approaches 1 is not equal to the function's value at x=1.

69 Easy

69 of 930

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x,x < 0}\\ {1,x = 0}\\ {{x^2},x > 0} \end{array}} \right.

, then

\lim\limits_{x \to 0} f\left( x \right)

View answer and explanation

Correct option: A

Option A (0) is correct because both the left-hand limit and the right-hand limit of f(x) as x approaches 0 are equal to 0.

70 Easy

70 of 930

Understanding +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}}\left[\frac{x}{\sin x}\right]

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

View answer and explanation

Correct option: A

Option A (1) is correct because for x near 0, x/sin(x) is greater than 1, and the greatest integer function of a number slightly greater than 1 is 1.

71 Easy

71 of 930

Application +4 | -1

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

View answer and explanation

Correct option: B

e^{-2}

The correct answer is B (

e^{-2}

) because it is the limit calculated using the standard limit

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

72 Medium

72 of 930

Understanding +4 | -1

The left hand limit of the function

f(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{{|x - 4|}}{{x - 4}},}&{x \ne 4}\\ {0,}&{x = 4} \end{array}} \right.

x=4

, is

View answer and explanation

Correct option: B

-1

Option A (Increasing on

(0, \pi / 2)

) is correct because the derivative of

f(x)

is positive on this interval, indicating that

f(x)

is monotonically increasing.

73 Easy

73 of 930

Analyzing +4 | -1

f(x)= \begin{cases}x^2, & \text { when } x \text { is an integer } \\ \frac{k\left(x^2-4\right)}{2-x}, & \text { otherwise }\end{cases}

Then,

\lim \limits_{{x \rightarrow 2}} f(x)

View answer and explanation

Correct option: B

Exists for every real

k

The correct answer is B (Exists for every real k) because the left-hand limit and right-hand limit are equal for all values of k.

74 Easy

74 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} {\left( { - 1} \right)^{\left[ x \right]}}

, where [.] denotes the greatest integer function 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 of the function as x approaches 0 are not equal.

75 Medium

75 of 930

Analyzing +4 | -1

f(x)

is differentiable and

f^{\prime}(0) \neq 0

, then the value of

\lim _{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because after applying L'Hôpital's rule, the limit simplifies to -1, given that f'(0) is not zero.

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76 Medium

76 of 930

Understanding +4 | -1

The value of

\lim\limits _{x \rightarrow k^{-}} x-[x]

, where

k

is an integer, is

View answer and explanation

Correct option: B

Option B (1) is correct because as x approaches k from the left, x - [x] approaches 1.

77 Easy

77 of 930

Application +4 | -1

Evaluate

\lim \limits_{{x \rightarrow 0}} \frac{(1-x)^n-1}{x}

View answer and explanation

Correct option: B

-n

Option B (-n) is correct because using the limit definition and binomial expansion, the limit evaluates to -n.

78 Easy

78 of 930

Application +4 | -1

f\left( 1 \right) = 1,\;f'\left( 1 \right) = 2

, then

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

View answer and explanation

Correct option: A

Option A (2) is correct because it is obtained by applying L'Hopital's rule to the limit and using the given values of f(1) and f'(1).

79 Easy

79 of 930

Analyzing +4 | -1

\lim \limits_{x \rightarrow 0^{-}}(-\ln (\{x\}+|[x]|))^{\{x\}}

is equal to

View answer and explanation

Correct option: D

\ln \frac{1}{2}

Option D (

\ln \frac{1}{2}

) is correct because the expression simplifies to

-\ln 2

as x approaches 0 from the left, which is equivalent to

\ln \frac{1}{2}

80 Easy

80 of 930

Analyzing +4 | -1

\lim _{x \rightarrow-1} \frac{1+\sqrt[n]{x}}{1+\sqrt[m]{x}}

(m,n are odd positive integers) is

View answer and explanation

Correct option: A

\frac{m}{n}

Option A (

\frac{m}{n}

) is correct because by substituting

x = z^{mn}

, the limit is transformed and evaluated using the formula for the limit of

\frac{x^n - a^n}{x-a}

81 Medium

81 of 930

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\sin x,x \ne n\mathrm{ }\pi ,n \in Z}\\ {0,\mathrm{other}\,\,\,\mathrm{wise}} \end{array}} \right.

and

g\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} + 1,x \ne 0,2}\\ {4,x = 0}\\ {5,x = 2} \end{array}} \right.

, then

\lim\limits_{x \to 0} g\left( {f\left( x \right)} \right)

View answer and explanation

Correct option: A

Option A (1) is correct because as x approaches 0, sin(x) also approaches 0, and g(0) = 4. However, since x != n*pi, we use the other definition of g(x) which is x² + 1. Thus, the limit is 0² + 1 = 1.

82 Easy

82 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because both the numerator and denominator approach π/2 as x approaches infinity.

83 Easy

83 of 930

Analyzing +4 | -1

Let

f: R \rightarrow R

be a positive increasing function with

\lim \limits_{{x \rightarrow \infty}} \frac{f(3 x)}{f(x)}=1

. Then,

\lim \limits_{{x \rightarrow \infty}} \frac{f(2 x)}{f(x)}=

View answer and explanation

Correct option: C

Option C (1) is correct because by using the Squeeze Theorem and the given condition

\lim \limits_{{x \rightarrow \infty}} \frac{f(3 x)}{f(x)}=1

, we can conclude that

\lim \limits_{{x \rightarrow \infty}} \frac{f(2 x)}{f(x)}=1

84 Easy

84 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to \sqrt 3 } \left[ {{x^2}} \right],

View answer and explanation

Correct option: C

Non-existent

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

85 Easy

85 of 930

Analyzing +4 | -1

\lim\limits_{x \to 1} {\left( {\log ex} \right)^{1/\log x}}

is equal to

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because by rewriting the expression and using the limit definition of

e

, the limit evaluates to

e

86 Easy

86 of 930

Understanding +4 | -1

\lim\limits_{n \to \infty } {r^n} = 0

, then

r

is equal to

View answer and explanation

Correct option: A

\frac{4}{5}

Option A (

\frac{4}{5}

) is correct because for

\lim\limits_{n \to \infty } {r^n} = 0

, the absolute value of

r

must be less than 1, i.e.,

|r| < 1

87 Easy

87 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}(1+x)^{\operatorname{cosec} x}

is equal to

View answer and explanation

Correct option: C

e

Option C (e) is correct because by using the limit property

\lim_{x \to 0} (1 + x)^{\frac{{1}}{{x}}} = e

, the limit evaluates to e.

88 Easy

88 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}

equals

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because the limit can be simplified using trigonometric identities and standard limits.

89 Easy

89 of 930

Understanding +4 | -1

The value of

\lim\limits_{x \to \infty } {\left\{ {\frac{{a_1^{1/x} + a_2^{1/x} + \ldots + a_n^{1/x}}}{n}} \right\}^1}

, is

View answer and explanation

Correct option: D

{a_1}{a_2} \ldots \ldots {a_n}

Option D (

{a_1}{a_2} \ldots \ldots {a_n}

) is correct because it uses the limit properties and exponential/logarithmic transformations to simplify the expression.

90 Easy

90 of 930

Analyzing +4 | -1

\lim\limits_{x \to 0} {\left( {\mathrm{cosec}x} \right)^{1/\log x}}

is equal to

View answer and explanation

Correct option: C

\frac{1}{e}

Option C (1/e) is correct because after applying logarithms and L'Hospital's rule, the limit evaluates to 1/e.

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