$\mathop {\lim }\limits_{x \to \pi /4} \,\,\frac{{\mathop \smallint \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - {\pi ^2}/16}}$ equals

$\frac{2}{\pi }f\left( 2 \right)$

$\frac{2}{\pi }f\left( {\frac{1}{2}} \right)$

Let $\alpha $ and $\beta $ be the distinct roots of $a{x^2} + bx + c = 0$, then $\mathop {\lim }\limits_{x \to \alpha } \frac{{1 - \cos \left( {a{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}$ is equal to

$\frac{1}{2}{\left( {\alpha - \beta } \right)^2}$

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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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Showing 91-120 of 267 questions

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91 Medium

91 of 267

Analyzing +4 | -1

If [x] denotes the greatest integer less than or equal to

x

, then the value of

\lim\limits_{x \to 1} \left\{ {1 - x + \left[ {x - 1} \right] + \left[ {1 - x} \right]} \right\}

View answer and explanation

Correct option: C

- 1

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

92 Medium

92 of 267

Application +4 | -1

The value of

\lim \limits_ {x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x}

View answer and explanation

Correct option: A

Option A (1) is correct because the limit can be evaluated by using L'Hôpital's Rule after transforming the expression into an exponential form.

93 Medium

93 of 267

Application +4 | -1

Let

f(x)= \begin{cases}1 /|x| & \text { for }|x| \geq 1 \\ a x^2+b & \text { for }|x|<1\end{cases}

The coefficients

a

and

b

so that

f

is continuous and differentiable at any point, are equal to

View answer and explanation

Correct option: A

a=-1 / 2, b=3 / 2

Option A (

a=-1 / 2, b=3 / 2

) is correct because these values ensure that the function is both continuous and differentiable at

x = 1

and

x = -1

94 Medium

94 of 267

Application +4 | -1

Let

f(x)

be a differentiable function at

x=a,

f^1(a)=4,

f(a)=2.

Then

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

is equal to

View answer and explanation

Correct option: B

2-4 a

Option B (

2-4a

) is correct because applying L'Hopital's rule and using the given values of

f(a)

and

f'(a)

leads to this result.

95 Medium

95 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{4+\sin 2 x+A \sin x+B \cos x}{x^2}

exists, then the values of

A

and

B

are

View answer and explanation

Correct option: A

-2

and

-4

The correct answer is A (-2 and -4) because for the limit to exist, both the numerator and its derivative must approach zero as x approaches zero, leading to A = -2 and B = -4.

96 Medium

96 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\log (\sin 7 x+\cos 7 x)}{\sin 3 x}

equals :

View answer and explanation

Correct option: B

\frac{7}{3}

Option B (

\frac{7}{3}

) is correct. By using L'Hopital's rule or series expansion, the limit evaluates to

\frac{7}{3}

97 Medium

97 of 267

Application +4 | -1

\Delta(x)=\left|\begin{array}{cc}e^x & \sin x \\ \cos x & \ln \left(1+x^2\right)\end{array}\right|

, then the value of

\lim _{x \rightarrow 0} \frac{\Delta(x)}{x}

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because by applying L'Hôpital's rule or using standard limits, the limit evaluates to -1.

98 Medium

98 of 267

Application +4 | -1

\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} \frac{f(x)-2}{2 x^2-1}=3 \pi

, then

\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} e^{f(x)}

will be

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit of f(x) as x approaches -1/sqrt(2) is 2, so the limit of e^(f(x)) is e².

99 Medium

99 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{\ln (a+x)-\ln a}{x}\right)+k \lim \limits_{{x \rightarrow e}} \frac{\ln x-1}{x-e}=1

then

View answer and explanation

Correct option: A

k=e\left(1-\frac{1}{a}\right)

Option A (

k=e\left(1-\frac{{1}}{{a}}\right)

) is correct because by applying the definition of derivative and simplifying, we can find the value of k.

100 Medium

100 of 267

Application +4 | -1

f

is a differentiable function such that

f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}, x \neq 0, y \neq 0, f(y) \neq 0

. If

f^{\prime}(1)=2

, then

f^{\prime}(x)=\ldots \ldots

View answer and explanation

Correct option: D

\frac{2 f(x)}{x}

Option C (

\frac{\pi}{3}

) is correct because by using the sine rule and differentiating, we find the angle to be

\frac{\pi}{3}

101 Medium

101 of 267

Application +4 | -1

Let

f(x)=\left\{\begin{array}{cl}e^{-\frac{1}{x^2}} \sin \frac{1}{x} & x \neq 0, \text { then } f^{\prime}(0)= \\ 0 & x=0\end{array}\right.

View answer and explanation

Correct option: C

Option C (0) is correct because the derivative at x=0 is found using the limit definition and the squeeze theorem.

102 Medium

102 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}

View answer and explanation

Correct option: C

Is equal to

\frac{27}{2}

Option C (Is equal to

\frac{27}{2}

) is correct because by applying binomial expansion and simplifying the limit, we get

\frac{27}{2}

103 Medium

103 of 267

Application +4 | -1

\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=\mathrm{e}^2

, then the values of

\mathrm{a} \& \mathrm{~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 can be evaluated using L'Hopital's rule or by comparing coefficients, leading to a=1 while b can be any real number.

104 Medium

104 of 267

Application +4 | -1

Which of the following is/are true?

I. The derivative of

x^2-2

x=10

is 18.

II. The derivative of

99 x

x=100

is 99.

III. The derivative of

x

x=1

is 1.

View answer and explanation

Correct option: C

II and III are true

Option C is correct because statements II and III are true. The derivative of 99x is 99, and the derivative of x is 1, while the derivative of x²-2 at x=10 is 20, not 18.

105 Medium

105 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{x \sqrt[3]{z^2-(z-x)^2}}{\left(\sqrt[3]{8 x z-4 x^2}+\sqrt[3]{8 x z}\right)^4}

is equal to

View answer and explanation

Correct option: B

\frac{1}{2^{23 / 3} \cdot z}

Option B (1/(2^(23/3) * z)) is correct because after simplifying the expression and taking the limit, the result is 1/(2^(23/3) * z).

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106 Medium

106 of 267

Analyzing +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}

is equal to

View answer and explanation

Correct option: B

The correct answer is B (0) because as n approaches infinity, the terms (2^(1/2) - 2^(1/(2n+1))) approach (2^(1/2) - 1), and since each term is less than 1, the infinite product converges to 0.

107 Medium

107 of 267

Application +4 | -1

\lim\limits_{x \to \pi /4} \,\,\frac{{\int \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - {\pi ^2}/16}}

equals

View answer and explanation

Correct option: A

\frac{8}{\pi }f\left( 2 \right)

Option A (

\frac{8}{\pi }f\left( 2 \right)

) is correct because applying Leibniz's rule and L'Hopital's rule results in this expression.

108 Medium

108 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{1^2 \cdot n+2^2 \cdot(n-1)+3^2 \cdot(n-2)+\ldots+n^2 \cdot 1}{1^3+2^3 \ldots+n^3}

, is equal to

View answer and explanation

Correct option: A

1 / 3

Option A (1/3) is correct because the limit is evaluated by simplifying the summation expressions and taking the limit as n approaches infinity.

109 Medium

109 of 267

Analyzing +4 | -1

f\left( x \right)

is differentiable function and

f''\left( 0 \right) = a

, then

\lim\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}

is equal to

View answer and explanation

Correct option: A

3a

Option A (

3a

) is the correct answer because it is obtained by applying L'Hopital's rule twice and then evaluating the limit.

110 Medium

110 of 267

Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{{e^{1/x}}}}{{{e^{1/x + 1}}}}

is equal to

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the limit depends on the direction of approach to 0, and the limit is 1/e.

111 Medium

111 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left\{1+x \ln \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta

b>0

and

\theta \in(-\pi, \pi]

, then the value of

\theta

, is

View answer and explanation

Correct option: D

\pm \frac{\pi}{2}

Option D (

\pm \frac{\pi}{2}

) is correct because by simplifying the given limit equation, we find that

\sin^2 \theta = 1

, which leads to

\theta = \pm \frac{\pi}{2}

112 Medium

112 of 267

Analyzing +4 | -1

\lim\limits_{{\text{x}} \to 0} \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}

is equal to:

View answer and explanation

Correct option: C

\frac{1}{6}

Option C (

\frac{1}{6}

) is correct because the limit is evaluated using trigonometric identities and series expansion.

113 Medium

113 of 267

Application +4 | -1

\lim\limits_{x \to 0} \,\frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}\,

is equal to

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because using trigonometric identities and limit properties, the given limit simplifies to

\pi

114 Medium

114 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{1+\sin x-\cos x+\log _e(1-x)}{x^3}

, is

View answer and explanation

Correct option: C

-1 / 2

Option C (-1/2) is correct because the limit is evaluated using the Taylor series expansions of

\sin x

\cos x

, and

\log_e(1-x)

115 Medium

115 of 267

Analyzing +4 | -1

\lim\limits_{x \to - 1} \frac{{\left( {1 + x} \right)\left( {1 - {x^2}} \right)\left( {1 + {x^3}} \right)\left( {1 - {x^4}} \right) \ldots \left( {1 - {x^{4n}}} \right)}}{{{{\left[ {\left( {1 + x} \right)\left( {1 - {x^2}} \right)\left( {1 + {x^3}} \right)\left( {1 - {x^4}} \right) \ldots \left( {1 - {x^{2n}}} \right)} \right]}^2}}}

is equal to

View answer and explanation

Correct option: A

^{4n}{C_{2n}}

Option A (

^{4n}{C_{2n}}

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

^{4n}{C_{2n}}

116 Medium

116 of 267

Application +4 | -1

Let

\alpha

and

\beta

be the distinct roots of

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

, then

\lim\limits_{x \to \alpha } \frac{{1 - \cos \left( {a{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}

is equal to

View answer and explanation

Correct option: D

\frac{{{a^2}}}{2}{\left( {\alpha - \beta } \right)^2}

Option D (

\frac{{{a^2}}}{2}{\left( {\alpha - \beta } \right)^2}

) is correct because by using trigonometric identities and the fact that alpha and beta are roots, the limit evaluates to the given expression.

117 Medium

117 of 267

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because after applying the series expansion of e^x, the limit evaluates to 1/2.

118 Medium

118 of 267

Application +4 | -1

\lim\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}}

is equal to

View answer and explanation

Correct option: B

{e^{ - 12}}

Option B (

{e^{ - 12}}

) is correct because the limit of the given expression as x approaches infinity equals

{e^{ - 12}}

after applying limit rules.

119 Medium

119 of 267

Application +4 | -1

f'\left( a \right) = 2

and

f\left( a \right) = 4

, then

\lim\limits_{x \to a} \frac{{xf\left( a \right) - af\left( x \right)}}{{x - a}}

equals

View answer and explanation

Correct option: B

4 - 2a

Option B (4 - 2a) is correct because it is the result of applying L'Hopital's rule or algebraic manipulation to the limit expression.

120 Medium

120 of 267

Application +4 | -1

Let

x_1=1 \quad \text { and } \quad x_{n+1}=\frac{4+3 x_n}{3+2 x_n} \text { for } n \geq 1

\lim _{n \rightarrow \infty} x_n

exists finitely, then the limit is equal to

View answer and explanation

Correct option: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct because the limit, if it exists, must satisfy the recursive relation, leading to the equation

l = \frac{4+3l}{3+2l}

, which solves to

l = \sqrt{2}

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