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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 121-150 of 647 questions

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121 Easy

121 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow e}} \frac{\log _e x-1}{|x-e|}

, is

View answer and explanation

Correct option: D

Non-existent

Option D (Non-existent) is correct because the left-hand limit and the right-hand limit are not equal, indicating that the limit does not exist.

122 Easy

122 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow a}}\left(2-\frac{a}{x}\right)^{\tan \frac{\pi x}{2 a}}

, is

View answer and explanation

Correct option: C

e^{-2 / \pi}

Option C (

e^{-2 / \pi}

) is correct because the limit can be evaluated using the exponential form of limits and L'Hopital's rule after substitution.

123 Easy

123 of 647

Application +4 | -1

Value of

\lim \limits_{{x \rightarrow 0}} \frac{\ell n \left(1+2 x-3 x^2+4 x^3\right)}{\ell n \left(1-5 x+6 x^2-7 x^3\right)}

is -

View answer and explanation

Correct option: C

-\frac{2}{5}

Option C (-2/5) is correct because the limit is evaluated using L'Hopital's rule or by manipulating the logarithmic expressions.

124 Easy

124 of 647

Understanding +4 | -1

The value of

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

View answer and explanation

Correct option: B

Option B (0) is correct because as x approaches 0 from the left,

\frac{{\sin x}}{{x}}

approaches 1 from below, and the greatest integer function of a value approaching 1 from below is 0.

125 Easy

125 of 647

Application +4 | -1

The derivative of function

f\left( x \right)

{\tan ^4}x

. If

f\left( x \right) = 0

, then

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

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because by applying L'Hopital's rule, the limit is equal to the limit of the derivative of the function, which is 0.

126 Easy

126 of 647

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

{e^{ - 2}}

Option C (

{e^{-2}}

) is correct because it is the result of manipulating the expression to fit the standard limit definition of e and then evaluating the limit.

127 Easy

127 of 647

Understanding +4 | -1

The value of

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

View answer and explanation

Correct option: C

Option C (1) is correct because as x approaches 0, the expression (

\frac{{x + 3}}{{x + 1}}

)

^x

approaches (

\frac{{3}}{{1}}

)

^0

, which equals 1.

128 Easy

128 of 647

Application +4 | -1

Let

f\left( x \right) = x\left[ x \right]

, where [x] denotes integral part of

x

. If

a

is not an integer, then

\lim\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} =

View answer and explanation

Correct option: C

[a]

Option C ([a]) is correct because since a is not an integer, for small h, [a+h] = [a], simplifying the limit calculation.

129 Easy

129 of 647

Understanding +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{1}{|1-x|}=

View answer and explanation

Correct option: D

\infty

Option D (

\infty

) is correct because as x approaches 1, the absolute value of (1-x) approaches 0, and the reciprocal approaches infinity.

130 Easy

130 of 647

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left( {\frac{1}{2}} \right)\left\{ {g\left( x \right) + \left( x \right)} \right\}\sin \left( x \right),x \ge 1}\\ {\sin x/x,\,\,\,x < 1} \end{array}} \right.

Where

g\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\mathrm{if}\,\,x > 0}\\ { - 1,\mathrm{if}\,\,x < 0}\\ {0,\mathrm{if}\,\,x = 0} \end{array}} \right.

Then,

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

is equal to

View answer and explanation

Correct option: C

\sin 1

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

131 Easy

131 of 647

Application +4 | -1

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

equals

View answer and explanation

Correct option: C

e^2

Option C (

e^2

) is correct because the limit can be evaluated using the standard limit

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

132 Easy

132 of 647

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}=

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 1/2.

133 Easy

133 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to - \pi } \frac{{\left| {x + \pi } \right|}}{{\sin x}}

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left-hand limit and the right-hand limit at x = -π are not equal, indicating that the limit does not exist.

134 Easy

134 of 647

Application +4 | -1

\alpha

is a repeated root of

a{x^2} + bx + c = 0

, then

\lim\limits_{x \to \alpha } \frac{{\sin \left( {a{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}

View answer and explanation

Correct option: B

a

Option B (

a

) is correct because the limit simplifies to

a

after substituting

a{x^2} + bx + c = a{\left( {x - \alpha } \right)^2}

and applying the limit.

135 Easy

135 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 3}} \frac{[x]^2-9}{x^2-9}

is equal to

View answer and explanation

Correct option: C

Does not exist

Option C (Does not exist) is correct because the left-hand limit and right-hand limit are not equal.

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136 Easy

136 of 647

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{5^x} - {5^{ - x}}}}{{2x}}

View answer and explanation

Correct option: A

\log 5

Option A (

\log 5

) is correct because the limit can be evaluated using L'Hopital's rule or by recognizing the derivative of an exponential function.

137 Easy

137 of 647

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^2 x \cot ^2(2 x)}

is equ

View answer and explanation

Correct option: D

Option D (1) is correct because after rewriting the expression using trigonometric identities and simplifying, the limit evaluates to 1.

138 Easy

138 of 647

Analyzing +4 | -1

If the normal to the curve

y=f(x)

x=0

be given by the equation

3 x-y+3=0

, then the value of

\lim \limits_{{x \rightarrow 0}} x^2\left\{f\left(x^2\right)-5 f\left(4 x^2\right)+4 f\left(7 x^2\right)\right\}^{-1}

, is

View answer and explanation

Correct option: B

-\frac{1}{3}

Option D (4, -4) is correct because it satisfies the given limit condition after applying L'Hopital's rule and using the information from the normal to the curve.

139 Easy

139 of 647

Application +4 | -1

Let

f

be a non-zero continuous function satisfying

f(x+y)=f(x) f(y)

for all

x, y \in \mathbf{R}

. If

f(2)=9

then

f(3)

View answer and explanation

Correct option: B

Option B (27) is correct because the function f(x) is an exponential function, and given f(2)=9, we can determine that f(x) = 3^x, thus f(3) = 27.

140 Easy

140 of 647

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because the limit can be rewritten in the form of the definition of

e

as n approaches infinity.

141 Easy

141 of 647

Analyzing +4 | -1

For

n \in N

, let

f_n(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)

(1+\sec 4 x) \ldots\left(1+\sec 2^n x\right)

. Then,

\lim \limits_{{x \rightarrow 0}} \frac{f_n(x)}{2 x}

View answer and explanation

Correct option: C

2^{n-1}

Option C,

2^{n-1}

, is correct because by repeatedly applying the identity

1 + \sec x = \frac{1 + \cos x}{\cos x} = \frac{2 \cos^2(x/2)}{\cos x}

, the function

f_n(x)

simplifies to

\tan(2^n x)

, and the limit can be evaluated using L'Hopital's rule or the standard limit

\lim_{x \to 0} \frac{\tan x}{x} = 1

142 Easy

142 of 647

Analyzing +4 | -1

\lim _{x \rightarrow 0}\left(\frac{3 x^2+2}{7 x^2+2}\right)^{1 / x^2}

is equal to:

View answer and explanation

Correct option: B

\frac{1}{e^2}

Option B (

\frac{1}{e^2}

) is correct because after simplifying the expression and evaluating the limit, the result is

\frac{1}{e^2}

143 Easy

143 of 647

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(2^{1 / 2} \cdot 2^{1 / 4} \cdot 2^{1 / 8} \ldots 2^{1 / 2^n}\right)

equals-

View answer and explanation

Correct option: B

2^1

Option B (

2^1

) is correct because the limit of the given expression simplifies to 2 raised to the power of 1, based on the sum of an infinite geometric series.

144 Easy

144 of 647

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1+2^4+3^4+\ldots+n^4}{n^5}-\lim \limits_{{n \rightarrow \infty}} \frac{1^3+2^3+3^3+\ldots+n^3}{n^5}

, is

View answer and explanation

Correct option: A

\frac{1}{5}

Option A (

\frac{1}{5}

) is correct because the difference of the two limits is equal to

\frac{1}{5}

145 Easy

145 of 647

Analyzing +4 | -1

The values of constants a and

b

so that

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

, are

View answer and explanation

Correct option: A

a=1, b=-\frac{3}{2}

Option A (

a=1, b=-\frac{3}{2}

) is correct because these values satisfy the given limit condition after simplifying the expression.

146 Easy

146 of 647

Understanding +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\left| x \right|}}{x}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand limit and the right-hand limit of |x|/x as x approaches 0 are not equal, hence the limit does not exist.

147 Easy

147 of 647

Application +4 | -1

f(x)= \begin{cases}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { for }-1 \leq x<0 \\ 2 x^2+3 x-2, & \text { for } 0 \leq x \leq 1\end{cases}

is continuous at

x=0

, then

k

is equal to

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because for the function to be continuous at x=0, the left-hand limit (LHL) must equal the right-hand limit (RHL).

148 Easy

148 of 647

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\int \nolimits_0^{{x^2}} \cos {t^2}dt}}{{x\sin x}}

View answer and explanation

Correct option: B

Option B (1) is correct because L'Hopital's rule can be applied twice to evaluate the limit, resulting in 1.

149 Easy

149 of 647

Application +4 | -1

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

, then

\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}

View answer and explanation

Correct option: B

\frac{53}{3}

Option B (53/3) is correct because it is obtained by applying L'Hopital's rule and using the given function and its derivative.

150 Easy

150 of 647

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because by applying L'Hopital's rule multiple times, the limit simplifies to

\frac{1}{2}

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