$\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$ 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 · 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 151-180 of 930 questions

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151 Easy

151 of 930

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.

152 Medium

152 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\left(\frac{1-\cos ^{2}(3 x)}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)

is equal to

View answer and explanation

Correct option: C

Option C (18) is correct because after applying L'Hopital's rule and trigonometric limits, the expression simplifies to 18.

153 Easy

153 of 930

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.

154 Medium

154 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{(1+x)^{y x}-e}{x}

View answer and explanation

Correct option: B

\frac{-e}{2}

Option B (-e/2) is correct because after applying L'Hôpital's rule and simplifying, the limit evaluates to -e/2.

155 Easy

155 of 930

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.

156 Easy

156 of 930

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.

157 Hard

157 of 930

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

158 Easy

158 of 930

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}

159 Medium

159 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e^{1-a}

Option B (

e^{1-a}

) is correct because the limit can be evaluated using L'Hopital's rule and properties of exponential functions.

160 Medium

160 of 930

Understanding +4 | -1

f

is an odd function and if

\lim\limits _{x \rightarrow 0} f(x)

exists, then

\lim\limits _{x \rightarrow 0} f(x)

View answer and explanation

Correct option: A

Option A (0) is correct because for an odd function, if the limit exists as x approaches 0, it must be 0.

161 Easy

161 of 930

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.

162 Easy

162 of 930

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.

163 Medium

163 of 930

Analyzing +4 | -1

If [.] denotes the greatest integer function, then

\lim \limits_{{x \rightarrow 0}} \frac{\sin \left[-\sec ^2 x\right]}{1+[\cos x]}

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit evaluates to

-\sin 2

, which is not among the given options.

164 Easy

164 of 930

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

165 Easy

165 of 930

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.

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166 Easy

166 of 930

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.

167 Medium

167 of 930

Analyzing +4 | -1

f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7

, then

\lim \limits_{{h \rightarrow 0}} \frac{f(1-h)-f(1)}{h^3+3 h}

is equal to

View answer and explanation

Correct option: A

\frac{53}{3}

Option A (

\frac{53}{3}

) is correct because by applying the definition of the derivative and evaluating

f'(1)

, the limit simplifies to

\frac{53}{3}

168 Easy

168 of 930

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

169 Easy

169 of 930

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}

170 Easy

170 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 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.

171 Easy

171 of 930

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)

172 Easy

172 of 930

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.

173 Easy

173 of 930

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.

174 Easy

174 of 930

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.

175 Medium

175 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{3^{2 x}-1}{2^{3 x}-1}

View answer and explanation

Correct option: A

\frac{\log 9}{\log 8}

Option A (

\frac{{\log 9}}{{\log 8}}

) is correct because by using the limit property

\lim_{x \to 0} \frac{{a^x - 1}}{x} = \log a

, the limit evaluates to

\frac{{\log 9}}{{\log 8}}

176 Easy

176 of 930

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.

177 Easy

177 of 930

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

178 Easy

178 of 930

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.

179 Easy

179 of 930

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}

180 Medium

180 of 930

Analyzing +4 | -1

[x]

denotes the greatest integer less than or equal to

x

, then

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

, is

View answer and explanation

Correct option: C

Non-existent

The correct answer is C (Non-existent) because the limit does not exist as the left-hand limit and right-hand limit are not equal.

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