$\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$ is equal to

${\log _e}\left( {\frac{b}{a}} \right)$

${\log _e}\left( {a + b} \right)$

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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 91-120 of 647 questions

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91 Easy

91 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{1-x^{-2 / 3}}{1-x^{-1 / 3}}

is equal

View answer and explanation

Correct option: A

Option A (2) is correct because it is the result of factoring and simplifying the limit expression as x approaches 1.

92 Easy

92 of 647

Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(\frac{\sqrt[n]{p}+\sqrt[n]{q}}{2}\right)^n, p, q > 0

equals

View answer and explanation

Correct option: B

\sqrt{p q}

Option B (

\sqrt{pq}

) is correct because the limit can be evaluated by converting it to an exponential form and using L'Hopital's rule or series expansion.

93 Easy

93 of 647

Analyzing +4 | -1

\lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}\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 are not equal.

94 Easy

94 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\int\limits_{0}^x t \sin (10 t) d t}{x}

is equal to

View answer and explanation

Correct option: D

The correct answer is D (0) because after applying L'Hopital's rule, the limit simplifies to 0.

95 Easy

95 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{3{x^3}}}

is equal to

L

, then the value of

(6L + 1)

View answer and explanation

Correct option: D

Option D (2) is correct. The limit is evaluated using series expansions of inverse trigonometric functions, and then substituting the result into the expression 6L + 1.

96 Easy

96 of 647

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{\sin \left| x \right|}}{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 not equal.

97 Easy

97 of 647

Application +4 | -1

If the value of

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

is 100 , then

n=

View answer and explanation

Correct option: B

-100

Option B (-100) is correct because applying the limit definition and solving for n gives n = -100.

98 Easy

98 of 647

Application +4 | -1

\lim\limits_{x \to 0} \,\,\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}

is equal to

View answer and explanation

Correct option: A

{\log _e}\left( {\frac{a}{b}} \right)

Option A (

{\log _e}\left( {\frac{a}{b}} \right)

) is correct because it is the result of applying L'Hopital's rule or by breaking the limit into standard logarithmic limits.

99 Easy

99 of 647

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{3}{2}

The correct answer is A (

\frac{3}{2}

) because applying L'Hôpital's rule to the limit yields the value

\frac{3}{2}

100 Easy

100 of 647

Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} - 2x + 1}}{{{x^2} - 4x + 2}}} \right)^x}

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.

101 Easy

101 of 647

Analyzing +4 | -1

Left hand limit and right hand limit of the function

\frac{x}{|x|+x^2}

x=0,

are

View answer and explanation

Correct option: D

-1, 1

Option D (-1, 1) is correct because it accurately calculates the left-hand limit and right-hand limit of the given function at x=0.

102 Easy

102 of 647

Understanding +4 | -1

\lim \limits_{{x \rightarrow 1}} \sin |||x|-2|-3|

is equal to

View answer and explanation

Correct option: A

\sin 2

Option A (

\sin 2

) is correct because the function is continuous, allowing direct substitution to find the limit, resulting in

\sin|||1|-2|-3| = \sin 2

103 Easy

103 of 647

Analyzing +4 | -1

f^{\prime}(2)=6

and

f^{\prime}(1)=4

, then

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

View answer and explanation

Correct option: D

is equal to 3

Option D (is equal to 3) is correct because the limit can be evaluated using L'Hopital's rule, resulting in the value 3.

104 Easy

104 of 647

Application +4 | -1

\lim\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2}

, then the values of

a

and

b

are

View answer and explanation

Correct option: B

a = 1,\;b \in R

Option B (

a = 1,\;b \in R

) is correct because the limit evaluates to

e^{2a}

, which must equal

e^2

, implying

a=1

and

b

can be any real number.

105 Easy

105 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\log \cos x}{x}=

View answer and explanation

Correct option: A

Option A (0) is correct because

\lim \limits_{{x \rightarrow 0}} \frac{\log \cos x}{x} = 0

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106 Easy

106 of 647

Analyzing +4 | -1

The value of

\lim\limits_{n \to \infty } \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + ... + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because the given series telescopes, and the limit as n approaches infinity is

\frac{1}{2}

107 Easy

107 of 647

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 as x approaches infinity, the numerator approaches

\cot^{-1}(0) = \frac{\pi}{2}

and the denominator approaches

\sec^{-1}(\infty) = \frac{\pi}{2}

, resulting in a limit of 1.

108 Easy

108 of 647

Application +4 | -1

f

be a function such that

f\left( 9 \right) = 9

and

f'\left( 9 \right) = 3

, then

\lim\limits_{x \to 9} \frac{{\sqrt {f\left( x \right)} - 3}}{{\sqrt x - 3}}

is equal to

View answer and explanation

Correct option: B

Option B (3) is correct because the limit simplifies to 3 after multiplying by the conjugate and using the given values of

f(9)

and

f'(9)

109 Easy

109 of 647

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin \left( {1 + \left[ x \right]} \right)}}{{\left[ x \right]}},\mathrm{for}\,\,\,\left[ x \right] \ne 0}\\ {0,\,\,\,\,\,\,\,\,\,\,\mathrm{for}\,\,\,\,\,\left[ x \right] = 0} \end{array}} \right.

Where

\left[ x \right]

denotes the greatest integer not exceeding

x

, then

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

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because as x approaches 0 from the left, [x] approaches -1, and the expression simplifies to sin(0)/-1 which equals 0.

110 Easy

110 of 647

Analyzing +4 | -1

Let

f(x)=\sqrt{x}(\sqrt{x+4}-\sqrt{x})

. Then

\lim _{x \rightarrow+\infty} f(x)

View answer and explanation

Correct option: A

exists and is equal to 2

Option A (exists and is equal to 2) is correct because after rationalizing and simplifying the expression, the limit as x approaches infinity is 2.

111 Easy

111 of 647

Application +4 | -1

Let

f: R \rightarrow R

be such that

f(1)=3

and

f^{\prime}(1)=6

.
Then,

\lim \limits_{{x \rightarrow 0}}\left\{\frac{f(1+x)}{f(1)}\right\}^{1 / x}

equals

View answer and explanation

Correct option: C

e^2

Option C,

e^2

, is correct because it is the result of recognizing the limit as an exponential limit and using the definition of the derivative.

112 Easy

112 of 647

Analyzing +4 | -1

The largest value of the non-negative integer a for which

\lim \limits_{{\alpha \rightarrow 1}}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}=\frac{1}{4}

, is

View answer and explanation

Correct option: A

Option A (2) is correct because solving the limit equation leads to a quadratic equation for 'a', and the largest non-negative integer solution is 2.

113 Easy

113 of 647

Analyzing +4 | -1

f(x)=0

is a quadratic equation such that

f(-\pi)=f(\pi)=0

and

f\left(\frac{\pi}{2}\right)=-\frac{3 \pi^2}{4}

, then

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

, is equal to

View answer and explanation

Correct option: C

2 \pi

Option C (

2 \pi

) is correct because after applying L'Hopital's rule and simplifying, the limit evaluates to

2 \pi

114 Easy

114 of 647

Analyzing +4 | -1

\lim _{x \rightarrow 0}\left[\frac{1}{x} \operatorname{Sin}^{-1}\left(\frac{2 x}{1+x^2}\right)\right]=

View answer and explanation

Correct option: C

Option C (2) is correct because after using the identity Sin⁻¹(2x/(1+x²)) = 2Tan⁻¹(x) and simplifying, the limit as x approaches 0 is 2.

115 Easy

115 of 647

Analyzing +4 | -1

\lim\limits_{x \to \infty } \frac{{\int \nolimits_0^{2x} x{e^{{x^2}}}dx}}{{{e^{4{x^2}}}}}

equals

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because evaluating the integral and applying L'Hopital's rule results in the answer

\frac{1}{2}

116 Easy

116 of 647

Analyzing +4 | -1

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

(where [.] denotes the greatest integer function)

View answer and explanation

Correct option: C

Equals 0

Option C (Equals 0) is correct because both the left-hand limit and the right-hand limit of the given expression as x approaches

\pi/2

are equal to 0.

117 Easy

117 of 647

Analyzing +4 | -1

Let

f:R \to R

be a differentiable function satisfying

f'(3) + f'(2) = 0.

then

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

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because applying L'Hopital's rule and using the given condition

f'(3) + f'(2) = 0

simplifies the limit to 1.

118 Easy

118 of 647

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 2 a}} \frac{\sqrt{x-2 a}+\sqrt{x}-\sqrt{2 a}}{\sqrt{x^2-4 a^2}}

, is

View answer and explanation

Correct option: B

\frac{1}{2 \sqrt{a}}

Option B (

\frac{1}{2 \sqrt{a}}

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

119 Easy

119 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{1+x}-1}{x}

is equal to

View answer and explanation

Correct option: A

1 / 2

Option A (1/2) is correct because rationalizing the numerator and simplifying the expression leads to a limit of 1/2 as x approaches 0.

120 Easy

120 of 647

Application +4 | -1

Let

f(x)=4

and

f^{\prime}(x)=4

. Then

\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}

is given by

View answer and explanation

Correct option: C

-4

Option C (-4) is correct because applying L'Hopital's rule gives the correct limit value.

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