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 · 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 61-90 of 647 questions

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61 Easy

61 of 647

Application +4 | -1

\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 2}}} \right)^x}

is equal to

View answer and explanation

Correct option: A

{e^4}

Option A (

{e^4}

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

e

62 Easy

62 of 647

Analyzing +4 | -1

The value of

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

is equal to

View answer and explanation

Correct option: B

-3

Option B (-3) is correct because the limit is evaluated using the Taylor series expansions of

e^x

\log(1+x)

, and

(1-x)^{-2}

63 Easy

63 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 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(\frac{a}{b})

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

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

64 Easy

64 of 647

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\sin x}}{{\sqrt {{x^2}} }}

, is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the limit does not exist as the left-hand limit is -1 and the right-hand limit is 1.

65 Easy

65 of 647

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{x \cos x+\sin x}{x^2+\tan x}

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because applying L'Hopital's rule yields a determinate form after the first application.

66 Easy

66 of 647

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

\frac{1}{4}

Option C (

\frac{1}{4}

) is correct because the limit of the given expression as n approaches infinity is equal to

\frac{1}{4}

67 Easy

67 of 647

Application +4 | -1

Value of

\lim \limits_{{x \rightarrow 0}} \frac{\sin 4 x}{\sin 2 x}

View answer and explanation

Correct option: A

Option A is correct because the limit evaluates to 2.

68 Easy

68 of 647

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \left( {\frac{1}{x}} \right),\;x \ne 0}\\ {0,\;x = 0} \end{array}} \right.

Then,

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

View answer and explanation

Correct option: C

Is equal to 0

Option A (

{a^2} + {b^2} + {c^2} \ge bc + ca + ab

) is correct because it is a direct result of the AM-GM inequality or can be derived from the fact that the sum of squares is always non-negative.

69 Easy

69 of 647

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}

equals

View answer and explanation

Correct option: A

\frac{10}{3}

Option A (

\frac{10}{3}

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

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

and the identity

1 - \cos 2x = 2 \sin^2 x

70 Easy

70 of 647

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

-\frac{1}{2 e}

Option C (-1/(2e)) is correct because it is obtained by using the Taylor series expansion of the logarithmic function and simplifying the limit.

71 Easy

71 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+9}-3}

, is

View answer and explanation

Correct option: A

Option A (3) is correct because rationalizing both the numerator and denominator and simplifying leads to a limit of 3 as x approaches 0.

72 Easy

72 of 647

Analyzing +4 | -1

\lim \limits_{{n \rightarrow\infty}}\left[2 \log (3 n)-\log \left(n^2+1\right)\right]

is equal to

View answer and explanation

Correct option: C

2 \log 3

Option C (

2 \log 3

) is correct because after simplifying the limit using logarithm properties and evaluating as n approaches infinity, the result is

2 \log 3

73 Easy

73 of 647

Understanding +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1}{n}\left\{1+e^{1 / n}+e^{2 / n}+e^{3 / n}+\ldots .+e^{\frac{n-1}{n}}\right\}

is equal to

View answer and explanation

Correct option: C

e-1

Option C (

e-1

) is correct because the limit can be expressed as a Riemann sum, which evaluates to the integral of e^x from 0 to 1.

74 Easy

74 of 647

Analyzing +4 | -1

I_1=\lim \limits_{{x \rightarrow \infty}}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x

and

I_2=\lim \limits_{{x \rightarrow 0}}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x

, then

\left(I_1, I_2\right)

View answer and explanation

Correct option: A

(0,0)

Option A ((0,0)) is correct because both limits

I_1

and

I_2

evaluate to 0.

75 Easy

75 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: B

3 / 2

Option B (

3 / 2

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

e^{x^2}

and

\cos x

or by applying L'Hopital's rule twice.

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76 Easy

76 of 647

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{\sqrt{3 x^2-1}+\sqrt{2 x^2-1}}{4 x+3}

, is

View answer and explanation

Correct option: B

\frac{1}{4(\sqrt{3}-\sqrt{2})}

Option B (1/3) is correct because simplifying the expression using the sum of squares formula leads to a limit of 1/3 as n approaches infinity.

77 Easy

77 of 647

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.

78 Easy

78 of 647

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}

79 Easy

79 of 647

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.

80 Easy

80 of 647

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}}

81 Easy

81 of 647

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}

82 Easy

82 of 647

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.

83 Easy

83 of 647

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.

84 Easy

84 of 647

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

85 Easy

85 of 647

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.

86 Easy

86 of 647

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.

87 Easy

87 of 647

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.

88 Easy

88 of 647

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.

89 Easy

89 of 647

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.

90 Easy

90 of 647

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.

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