$\lim _\limits{x \rightarrow 0}\left(x^n \log _e x\right)$ is

does not exists either finitely or infinitely

If $x$ is a real number in [0, 1], then the value of $\mathop {\lim }\limits_{m \to \infty } \mathop {\lim }\limits_{n \to \infty } \left[ {1 + {{\cos }^{2m}}\left( {n!\pi {\rm{ }}x} \right)} \right]$ is given by

1 or 2 according as $x$ is rational or irrational

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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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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 267 questions

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61 Medium

61 of 267

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.

62 Medium

62 of 267

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}

63 Medium

63 of 267

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}}

64 Medium

64 of 267

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.

65 Medium

65 of 267

Application +4 | -1

The value of

\lim\limits_{x \to {0^ + }} {x^m}{\left( {\log x} \right)^n},m,n.\mathrm{ }N

View answer and explanation

Correct option: A

Option A (0) is correct because by applying L'Hopital's rule repeatedly, the limit evaluates to 0 as x approaches 0 from the positive side.

66 Medium

66 of 267

Analyzing +4 | -1

f(x)=\left\{\begin{array}{cl}\frac{[x]^2+\sin [x]}{[x]} & \text { for }[x] \neq 0 \\ 0 & \text { for }[x]=0\end{array}\right.

where

[x]

denotes the greatest integer function, then

\lim \limits_{{x \rightarrow 0}} f(x)

, is

View answer and explanation

Correct option: D

Non-existent

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

67 Medium

67 of 267

Analyzing +4 | -1

Let

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

and

f(x)=x^2 g(x)

for all

x, y \in R

, where

g(x)

is continuous function. Then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: D

Option D is correct because using the given functional equation and the definition of the derivative, we find that f'(x) = 0.

68 Medium

68 of 267

Application +4 | -1

Which of the following is/are true?

I. The derivative of

{f}({x})={x}^3

{x}^2

II. The derivative of

f(x)=\frac{1}{x^3}

\frac{-1}{x^2}

View answer and explanation

Correct option: D

Both I and II are false

Option D is correct because both statements are false. The derivative of x³ is 3x², and the derivative of 1/x³ is -3/x⁴.

69 Medium

69 of 267

Analyzing +4 | -1

\underset{x \rightarrow a-}{L t}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)(a>0)

when

[x]

is G.I.F. is

View answer and explanation

Correct option: D

Option D (0) is correct because as x approaches a from the left, the greatest integer function evaluates to 0, making the entire expression equal to 0.

70 Medium

70 of 267

Analyzing +4 | -1

The values of

a

and

b

such that

\lim\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 0

, are

View answer and explanation

Correct option: C

- \frac{5}{2}, - \frac{3}{2}

Option C (

\frac{{ - 5}}{2}, - \frac{3}{2}

) is correct because by expanding the trigonometric functions using Taylor series and equating coefficients, the values of a and b can be determined.

71 Medium

71 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}} \frac{\int\limits_{0}^{x^2} \cos t^2 d t}{x \sin x}

View answer and explanation

Correct option: B

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

72 Medium

72 of 267

Application +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

\frac{2}{\pi }

Option C (

\frac{2}{\pi }

) is correct because the limit is evaluated using L'Hopital's rule or by rewriting the expression using trigonometric identities and then applying standard limits.

73 Medium

73 of 267

Application +4 | -1

The value of

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

View answer and explanation

Correct option: C

\frac{3}{2}

Option D (

\frac{3}{2}

) is correct because by factoring and using trigonometric limit properties, the limit evaluates to 3/2.

74 Medium

74 of 267

Analyzing +4 | -1

Let

P_n=\frac{2^3-1}{2^3+1} \cdot \frac{3^3-1}{3^3+1}+\cdots+\frac{n^3-1}{n^3+1} ; n=2,3,4, \ldots

Then

\lim \limits_{{n \rightarrow \infty}} P_n

is equal to

View answer and explanation

Correct option: D

\frac{2}{3}

Option D (

\frac{2}{3}

) is correct because after factoring and telescoping the product, the limit as n approaches infinity is

\frac{2}{3}

75 Medium

75 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(x^n \log _e x\right)

View answer and explanation

Correct option: C

The correct answer is C (0) because rewriting the expression and applying L'Hôpital's rule yields the value 0.

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76 Medium

76 of 267

Application +4 | -1

Let

(\tan \alpha) x+(\sin \alpha) y=\alpha \cdot

and

(\alpha \cdot \operatorname{cosec} \alpha) x+(\cos \alpha) y=1

be two variable straight lines,

\alpha

, being the parameter. Let

\mathrm{P}

be the point of intersection of the lines. Find the limiting position of

\mathrm{P}

when

\alpha \rightarrow 0

View answer and explanation

Correct option: A

(2,-1)

Option A ((2,-1)) is correct because it represents the limiting position of the intersection point P as alpha approaches 0, found by evaluating the limits of x and y coordinates.

77 Medium

77 of 267

Analyzing +4 | -1

f(x)=\lim \limits_{{n \rightarrow \infty}} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 x\right)^{2 n}+5}

, then the set of values of

x

for which

f(x)=0

, is

View answer and explanation

Correct option: A

|2 x|>\sqrt{3}

Option A (

|2x| > \sqrt{3}

) is correct because the function f(x) equals zero if and only if

(\frac{3}{\pi} \tan^{-1} 2x)^2 > 1

, leading to

|2x| > \sqrt{3}

78 Medium

78 of 267

Analyzing +4 | -1

Let

f(x)=\frac{x-\{x+1\}}{x-\{x+2\}}

; where

\{x\}

is the fractional part of

x

, then

\lim _{x \rightarrow 1 / 3} f(x)

View answer and explanation

Correct option: B

has value 1

Option B (has value 1) is correct because as x approaches 1/3, both the numerator and denominator of the function approach zero, and after simplification, the limit equals 1.

79 Medium

79 of 267

Application +4 | -1

The value of

\lim \limits_{x \rightarrow 0^{+}}(\sin x)^{1 / \ell n x}

is-

View answer and explanation

Correct option: D

e

Option D (e) is correct because the limit is evaluated by taking the natural logarithm and applying L'Hopital's rule.

80 Medium

80 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}(1+a \sin x)^{\operatorname{cosec} x}=3

, then

a

View answer and explanation

Correct option: B

\ln 3

Option B (

\ln 3

) is correct because evaluating the limit using the exponential form gives

e^a = 3

, thus

a = \ln 3

81 Medium

81 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{a \sin x-b x+c x^2+x^3}{2 x^2 \log (1+x)-2 x^3+x^4}

exists and finite, then which of the following is not correct?

View answer and explanation

Correct option: A

a=6

Option A (

a=6

) is correct because the limit exists and is finite only if

a=6

b=6

, and

c=0

. Therefore,

a=6

is the correct value.

82 Medium

82 of 267

Analyzing +4 | -1

x

is a real number in [0, 1], then the value of

\lim\limits_{m \to \infty } \lim\limits_{n \to \infty } \left[ {1 + {{\cos }^{2m}}\left( {n!\pi \mathrm{ }x} \right)} \right]

is given by

View answer and explanation

Correct option: A

2 or 1 according as

x

is rational or irrational

Option A (2 or 1 according as

x

is rational or irrational) is correct. When x is rational, the limit is 2; when x is irrational, the limit is 1.

83 Medium

83 of 267

Analyzing +4 | -1

For each

x \in R

, let

[x]

be the greatest integer less than or equal to

x

. Then,

\lim \limits_ {x \rightarrow 1^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}

is equal to

View answer and explanation

Correct option: A

-\sin 1

Option A (-

\sin 1

) is correct because after substituting the values for x approaching 1 from the left and simplifying, the limit evaluates to -

\sin 1

84 Medium

84 of 267

Analyzing +4 | -1

Let

g\left( x \right) = \frac{{{{\tan }^3}x - \tan x}}{{\cos \left[ {x + \left( {\pi /4} \right)} \right]}}

then

\lim\limits_{x \to \frac{\pi }{4}} g\left( x \right)

View answer and explanation

Correct option: C

-4

Option C (-4) is correct because after simplifying the trigonometric expression and taking the limit as x approaches pi/4, the result is -4.

85 Medium

85 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{[(a-n) n x-\tan x] \sin n x}{x^2}=0

then

a

is equal to

View answer and explanation

Correct option: D

n+\frac{1}{n}

Option D (

n+\frac{1}{n}

) is correct because by using the limit condition and applying L'Hôpital's rule, the value of 'a' is found to be

n+\frac{1}{n}

86 Medium

86 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^{2013}}{e^{3 x}}+\left(\cos \frac{2}{x}\right)^{x^2}\right)

View answer and explanation

Correct option: D

{e}^{-2}

Option D (

{e}^{-2}

) is correct because the limit of the first term is 0 and the limit of the second term is

{e}^{-2}

87 Medium

87 of 267

Application +4 | -1

f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2

, then the value of

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

View answer and explanation

Correct option: A

Option A (5) is correct because the limit can be evaluated by recognizing it as the derivative of a product and applying the given values.

88 Medium

88 of 267

Application +4 | -1

The derivative of

\cos \left(x^2+1\right)

by using first principle is

View answer and explanation

Correct option: B

-2 x \sin \left(x^2+1\right)

Option B (

-2 x \sin \left(x^2+1\right)

) is correct because it is the derivative of

\cos \left(x^2+1\right)

obtained using the first principle of differentiation and trigonometric identities.

89 Medium

89 of 267

Analyzing +4 | -1

Let

f(x)=\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^2}

for

x \neq \pi / 6

. Then

\lim _{x \rightarrow \pi / 6} f(x)

View answer and explanation

Correct option: B

\frac{1}{36}

Option B (

\frac{1}{36}

) is correct. By using trigonometric identities and L'Hopital's rule, the limit can be evaluated to

\frac{1}{36}

90 Medium

90 of 267

Analyzing +4 | -1

\lim\limits_{x \to 0} x\log \sin x

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because by rewriting the expression and applying L'Hopital's rule twice, the limit evaluates to 0.

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