If $f(x) = \,\left\{ {\begin{array}{*{20}{c}} {\frac{{\sin (p + 1)x + \sin x}}{x},}&{x 0} \end{array}} \right.$ is continuous at $x = 0$ then the ordered pair $(p,q)$ is equal to

$\left( { - \frac{1}{2},\frac{3}{2}} \right)$

$\left( {\frac{5}{2},\frac{1}{2}} \right)$

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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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Showing 121-150 of 930 questions

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121 Medium

121 of 930

Analyzing +4 | -1

The value of

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

View answer and explanation

Correct option: A

\frac{11 e}{24}

Option A (

\frac{11 e}{24}

) is correct because the limit is evaluated using the series expansion of

(1+x)^{1/x}

and simplifying the expression.

122 Medium

122 of 930

Analyzing +4 | -1

Let

L = \lim\limits_{x \to 0} \,\,\frac{{a - \sqrt {{a^2} - {x^2}} - {x^2}/4}}{{{x^4}}},a > 0

. If

L

is finite, then

View answer and explanation

Correct option: A

a = 2,L = \frac{1}{{64}}

Option A (

a = 2, L = \frac{1}{{64}}

) is correct because these values make the limit finite after using the binomial expansion.

123 Easy

123 of 930

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{1}{n^2}(1+2+\cdots+n)

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because the sum of the first n natural numbers is n(n+1)/2, and dividing by n² and taking the limit as n approaches infinity gives 1/2.

124 Medium

124 of 930

Application +4 | -1

The value of

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

View answer and explanation

Correct option: C

e^{-2}

Option C (

e^{-2}

) is correct because the limit can be evaluated by rewriting the expression in the form of

e^{\lim_{x \rightarrow \infty} f(x)}

125 Easy

125 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} n \cos \frac{\pi}{4 n} \sin \frac{\pi}{4 n}=k

then

k

View answer and explanation

Correct option: A

\frac{\pi}{4}

Option A (

\frac{\pi}{4}

) is correct because evaluating the limit using trigonometric identities and limit properties results in

\frac{\pi}{4}

126 Easy

126 of 930

Analyzing +4 | -1

\lim\limits_{x \to 2} = \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}

is equal to

View answer and explanation

Correct option: A

\frac{1}{{8\sqrt 3 }}

Option A (

\frac{1}{{8\sqrt 3 }}

) is correct because after rationalizing and applying the limit, the expression evaluates to this value.

127 Easy

127 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x - 0} \frac{1}{{{x^3}}}\int \nolimits_0^x \frac{{t\log \left( {1 + t} \right)}}{{{t^4} + 4}}dt

View answer and explanation

Correct option: B

\frac{1}{{12}}

Option B (

\frac{1}{{12}}

) is correct because using L'Hopital's rule, the limit simplifies to

\frac{1}{{12}}

128 Easy

128 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}

, is

View answer and explanation

Correct option: C

\sqrt 2

Option C (

\sqrt{2}

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

\sqrt{2}

129 Easy

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

, is

View answer and explanation

Correct option: B

\frac{53}{3}

Option B (53/3) is correct because it is the result of applying L'Hopital's rule or recognizing the limit as a derivative and evaluating it.

130 Easy

130 of 930

Application +4 | -1

f(x)=\left(\frac{x^2+5 x+3}{x^2+x+2}\right)^x

then

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

View answer and explanation

Correct option: D

e^4

Option D (

e^4

) is correct because the limit is evaluated by rewriting the expression in the form of

e^x

and using the limit properties.

131 Easy

131 of 930

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e^{-12}

Option B (

e^{-12}

) is correct because the limit can be evaluated by rewriting the expression and using the standard limit

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

132 Easy

132 of 930

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 1}} \pi(1-x) \tan \left(\frac{\pi x}{2}\right)

must be

View answer and explanation

Correct option: A

Option A (2) is correct because by substituting x = 1 + h and using trigonometric identities and limits, the value is found to be 2.

133 Medium

133 of 930

Application +4 | -1

Let

f

is twice differentiable on

R

such that

f(0)=1, f^{\prime}(0)=0

and

f^{\prime \prime}(0)=-1

, then for a

\in R

\lim _{x \rightarrow \infty}\left(f\left(\frac{a}{\sqrt{x}}\right)\right)^x=

View answer and explanation

Correct option: C

e^{\frac{a^2}{2}}

Option C (

e^{\frac{a^2}{2}}

) is correct because it is the limit calculated using L'Hôpital's rule and the given conditions.

134 Easy

134 of 930

Analyzing +4 | -1

\lim\limits_{x \to \frac{\pi }{2}} {\tan ^2}x\left( {{{\left( {2{{\sin }^2}x + 3\sin x + 4} \right)}^{\frac{1}{2}}} - {{\left( {{{\sin }^2}x + 6\sin x + 2} \right)}^{\frac{1}{2}}}} \right)

is equal to

View answer and explanation

Correct option: A

01-Dec

Option C (-0.083333333) is incorrect. Option A (1/12) is correct because by applying L'Hôpital's rule or trigonometric identities, the limit can be evaluated to 1/12.

135 Medium

135 of 930

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \left( {\frac{{\int \nolimits_0^{{x^2}} {{\sec }^2}t\mathrm{ }dt}}{{x\sin x}}} \right)

View answer and explanation

Correct option: C

Option C (1) is correct because after applying L'Hopital's rule, the limit evaluates to 1.

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136 Easy

136 of 930

Analyzing +4 | -1

\lim\limits_{x \to 1} \frac{{{x^8} - 2x + 1}}{{{x^4} - 2x + 1}}

equals

View answer and explanation

Correct option: A

Option A (3) is correct because applying L'Hôpital's rule once gives the correct limit value of 3.

137 Easy

137 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\pi

Option A (

\pi

) is correct because the expression simplifies to

\pi

after applying trigonometric identities and limit properties.

138 Medium

138 of 930

Analyzing +4 | -1

\alpha ,\,\,\beta

are the distinct roots of

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

then,

\lim\limits_{x \to \beta } \frac{{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)}}{{{{(x \to \beta )}^2}}}

is equal to

View answer and explanation

Correct option: C

2\left( {{b^2} - 4c} \right)

Option C (

2\left( {{{b^2} - 4c}} \right)

) is correct because by factoring the quadratic and using L'Hopital's rule, the limit simplifies to

2(\beta - \alpha)^2

, which equals

2(b^2 - 4c)

139 Medium

139 of 930

Analyzing +4 | -1

\lim _{x \rightarrow \infty}\left[x \sqrt{x^2+4}-\sqrt{x^4+16}\right]=

View answer and explanation

Correct option: C

Option C (2) is correct because after simplifying the expression and taking the limit as x approaches infinity, the result is 2.

140 Medium

140 of 930

Analyzing +4 | -1

The value of

\lim\limits_{\alpha \to 0} \frac{{\left[ {\mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\sec \alpha } \right) + {{\cot }^{ - 1}}\left( {\tan \alpha } \right) + {{\cot }^{ - 1}}\cos \left( {{{\sin }^{ - 1}}\alpha } \right)} \right]}}{\alpha }

View answer and explanation

Correct option: C

- 2

Option C (

-2

) is correct because after simplifying the trigonometric inverse functions and applying L'Hopital's rule, the limit evaluates to -2.

141 Easy

141 of 930

Analyzing +4 | -1

f(x) = \,\left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin (p + 1)x + \sin x}}{x},}&{x < 0}\\ {q,}&{x = 0}\\ {\frac{{\sqrt {x + {x^2}} }}{{{x^{3/2}}}},}&{x > 0} \end{array}} \right.

is continuous at

x = 0

then the ordered pair

(p,q)

is equal to

View answer and explanation

Correct option: D

\left( { - \frac{3}{2},\frac{1}{2}} \right)

Option D

\left( { - \frac{3}{2},\frac{1}{2}} \right)

is correct because the function is continuous at x=0 only if the left-hand limit, right-hand limit, and the function's value at x=0 are equal.

142 Medium

142 of 930

Analyzing +4 | -1

[x]

denotes the greatest integer less than or equal to

x

, then,

\lim \limits_{{n \rightarrow \infty}} \frac{\left[1^2 x\right]+\left[2^2 x\right]+\left[3^2 x\right]+\ldots . .+\left[n^2 x\right]}{n^3}

equals

View answer and explanation

Correct option: B

x / 3

Option B (

x / 3

) is correct because the limit is evaluated using the Squeeze Theorem and the sum of squares formula.

143 Medium

143 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (-

\frac{1}{2}

) is correct because the limit is evaluated using L'Hopital's rule or series expansion.

144 Easy

144 of 930

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

-\frac{e}{2}

Option C (

-\frac{e}{2}

) is correct because the limit is evaluated using the series expansion of

(1+x)^{1/x}

and simplifying the expression.

145 Medium

145 of 930

Analyzing +4 | -1

Let

f\left( x \right)

be twice differentiable function such that

f''\left( 0 \right) = 2

. Then,

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

View answer and explanation

Correct option: A

Option A (6) is correct because by applying L'Hôpital's Rule twice and using the given condition

f''(0) = 2

, the limit evaluates to 6.

146 Medium

146 of 930

Analyzing +4 | -1

\underset{x\to 1}{\mathop{Lt}}\,\left[ {{\left[ \frac{4}{{{x}^{2}}-{{x}^{-1}}}-\frac{1-3x+{{x}^{2}}}{1-{{x}^{3}}} \right]}^{-1}}+\frac{3\left( {{x}^{4}}-1 \right)}{{{x}^{3}}-{{x}^{-1}}} \right]=

View answer and explanation

Correct option: B

Option B (3) is correct because after simplification and applying the limit, the expression evaluates to 3.

147 Easy

147 of 930

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.

148 Easy

148 of 930

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.

149 Easy

149 of 930

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.

150 Easy

150 of 930

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.

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