If ${A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\;i = 1,\;2,\; \ldots ,\;n$ and if ${a_1} < {a_2} < {a_3} < \ldots < {a_n}$. Then, $\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2} \ldots {A_n}} \right),\;1 \le m \le n$

Is equal to ${\left( { - 1} \right)^m}$

Is equal to ${\left( { - 1} \right)^{m + 1}}$

Is equal to ${\left( { - 1} \right)^{m - 1}}$

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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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Showing 151-180 of 267 questions

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151 Medium

151 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left(\sin ^4 x+\frac{1}{4} \sin ^4 2 x+\ldots \cdot \frac{1}{4^n} \sin ^4\left(2^n x\right)\right)

View answer and explanation

Correct option: A

\sin ^2 x

The correct answer is

\sin^2 x

because the limit of the given series telescopes to

\sin^2 x

as n approaches infinity.

152 Medium

152 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow a}} \frac{x^a-a^x}{x^x-a^a}

equals

View answer and explanation

Correct option: A

\frac{1-\log a}{1+\log a}

Option A ((1-log a)/(1+log a)) is correct because after applying L'Hôpital's rule, the limit simplifies to (1-log a)/(1+log a).

153 Medium

153 of 267

Analyzing +4 | -1

{A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\;i = 1,\;2,\; \ldots ,\;n

and if

{a_1} < {a_2} < {a_3} < \ldots < {a_n}

. Then,

\lim\limits_{x \to {a_m}} \left( {{A_1}{A_2} \ldots {A_n}} \right),\;1 \le m \le n

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 of the product of A_i's as x approaches a_m are not equal, thus the limit does not exist.

154 Medium

154 of 267

Analyzing +4 | -1

The value of

\lim\limits_{x \to 2} \frac{5}{{\sqrt 2 - \sqrt x }},

View answer and explanation

Correct option: D

Non-existent

Option D (Non-existent) is correct because the left-hand limit approaches infinity and the right-hand limit approaches negative infinity.

155 Medium

155 of 267

Analyzing +4 | -1

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}

equals :

View answer and explanation

Correct option: C

\frac{1}{16}

Option A (

-\sin 1

) is correct because after substituting and simplifying, the limit equals

-\sin 1

156 Medium

156 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sin x+\log (1-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 expansions of sin(x) and log(1-x), the limit evaluates to -1/2.

157 Medium

157 of 267

Analyzing +4 | -1

\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because applying L'Hopital's rule to the limit results in the expression

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

158 Medium

158 of 267

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} 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 are different, hence the limit does not exist.

159 Medium

159 of 267

Application +4 | -1

f^{\prime}(2)=1

, then

\lim \limits_{{h \rightarrow 0}} \frac{f\left(2+h^2\right)-f\left(2-h^2\right)}{2 h^2}=\ldots \ldots

View answer and explanation

Correct option: B

Option B (1) is correct because by applying L'Hopital's rule, the limit simplifies to f'(2), which is given as 1.

160 Medium

160 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}}\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right) \ldots\left(1+\frac{1}{a_n}\right)

, where

a_1=1

and

a_n=n\left(1+a_{n-1}\right), \forall n \geq 2

, is

View answer and explanation

Correct option: A

e

Option A (e) is correct because the given limit simplifies to the series expansion of e after using the recurrence relation and taking the limit as n approaches infinity.

161 Medium

161 of 267

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: A

\frac{3}{{\sqrt 2 }}

Option A (

\frac{3}{{\sqrt 2 }}

) is correct because it is obtained by using a substitution and then evaluating the limit.

162 Medium

162 of 267

Analyzing +4 | -1

Evaluate

\lim \limits_{{x \rightarrow 1}} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}

View answer and explanation

Correct option: B

Option B (3) is correct because the limit is evaluated by factoring and simplifying the expression.

163 Medium

163 of 267

Analyzing +4 | -1

The value of

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

\left.\ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^n

View answer and explanation

Correct option: B

e^2

Option B (

e^2

) is correct because the given limit can be simplified to an infinite product of exponentials, which converges to

e^2

after summing the exponents.

164 Medium

164 of 267

Analyzing +4 | -1

The value of

\lim\limits_{x \to 2} \left\{ {{{\left[ {\frac{{{x^3} - 4x}}{{{x^3} - 8}}} \right)}^{ - 1}} - {{\left( {\frac{{x + \sqrt {2x} }}{{x - 2}} - \frac{{\sqrt 2 }}{{\sqrt x - \sqrt 2 }}} \right)}^{ - 1}}} \right\}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because after simplification and applying the limit, the expression evaluates to

\frac{1}{2}

165 Medium

165 of 267

Understanding +4 | -1

Let

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

Greatest integer less than or equal to

x

and

k

be an integer. Then, which one of the following is not correct?

View answer and explanation

Correct option: C

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

exists

Option C (

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

exists) is correct because the limit does not exist as the left-hand limit and right-hand limit are not equal.

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166 Medium

166 of 267

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because by applying L'Hopital's rule, the limit can be evaluated to 2.

167 Medium

167 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{1-\cos x}{e^{a x}-b x-1}=2

then the value of

\frac{a}{b}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because applying L'Hopital's rule twice and using the given limit, we find that a must equal b, hence a/b = 1.

168 Medium

168 of 267

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: B

\frac{{ - 1}}{2}

Option B (

\frac{{ - 1}}{2}

) is correct because after expanding the trigonometric and logarithmic functions using their Maclaurin series and simplifying, the limit evaluates to -1/2.

169 Medium

169 of 267

Application +4 | -1

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

is equal to-

View answer and explanation

Correct option: B

2 \ell n 3

Option B (2 ln 3) is correct because the limit is evaluated by rationalizing the denominator and using the exponential limit.

170 Medium

170 of 267

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \{ {\log _{\left( {n - 1} \right)}}n.{\log _n}\left( {n + 1} \right).{\log _{\left( {n + 1} \right)}}\left( {n + 2} \right) \ldots \left\{ { \ldots {{\log }_{\left( {{n^k} - 1} \right)}}\left( {{n^k}} \right)} \right\}

, is

View answer and explanation

Correct option: C

k

Option C (

k

) is correct because after applying the change of base formula for logarithms and L'Hopital's rule, the limit evaluates to

k

171 Medium

171 of 267

Analyzing +4 | -1

\lim\limits_{\theta \to 0} \frac{{4\theta \left( {\tan \theta - 2\theta \tan \theta } \right)}}{{\left( {1 - \cos 2\theta } \right)}}

View answer and explanation

Correct option: D

Option D (2) is correct because applying L'Hopital's rule twice to the limit results in a value of 2.

172 Medium

172 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 5}} \frac{x^k-5^k}{x-5}=500

, then

k

is equal to:

View answer and explanation

Correct option: A

Option A (4) is correct because the value of k is 4, derived from the limit definition of the derivative.

173 Medium

173 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow \infty}} \frac{\int\limits_{0}^x \tan ^{-1} t \, d t}{\sqrt{x^2+1}}

is equal to:

View answer and explanation

Correct option: A

\frac{\pi}{2}

Option A (

\frac{\pi}{2}

) is correct because after applying L'Hôpital's rule, the limit simplifies to

\frac{\pi}{2}

174 Medium

174 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow 2 a}} \frac{\sqrt{x-2 a}+\sqrt{x}-\sqrt{2 a}}{\sqrt{x^2-4 a^2}}

View answer and explanation

Correct option: B

\frac{1}{2 \sqrt{a}}

Option B (

\frac{1}{2 \sqrt{a}}

) is correct because after simplifying and rationalizing, the limit evaluates to

\frac{1}{2 \sqrt{a}}

175 Medium

175 of 267

Application +4 | -1

The value of

\operatorname{Lim}_{x \rightarrow 0} \frac{\int_0^x x e^{t^2} d t}{1+x-e^x}

is equal to

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because applying L'Hôpital's Rule twice to the limit

\operatorname{Lim}_{x \rightarrow 0} \frac{\int_0^x x e^{t^2} d t}{1+x-e^x}

yields -2.

176 Medium

176 of 267

Application +4 | -1

[.]

denotes the greatest integer function, then

\lim \limits_{{x \rightarrow \pi / 2}}\left[\frac{x-\frac{\pi}{2}}{\cos x}\right]

is equal to

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because by substituting

x = \frac{\pi}{2} + \theta

and using the limit

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

, the expression simplifies to -2.

177 Medium

177 of 267

Application +4 | -1

A non-zero function

f

satisfies

f(x-y)=\frac{f(x)}{f(y)}, \forall x, y \in \mathbf{R}

. If

f(0)=5

and

f(5)=5

, then

f(5)=\ldots \ldots

View answer and explanation

Correct option: A

Option A (1) is correct because by substituting x=5 and y=0 into the given functional equation, we can derive that f(5) = 1.

178 Medium

178 of 267

Understanding +4 | -1

Assertion:

\lim \limits_{{x \rightarrow 0}}(1+3 x)^{1 / x}=e^3

.

Reason: Since

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

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct
explanation for assertion

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the limit can be evaluated using the given standard limit.

179 Medium

179 of 267

Analyzing +4 | -1

In a circle of radius

r

, an isosceles

\Delta A B C

is inscribed with

A B=A C

. If

\Delta A B C

has perimeter

p=2\left[\sqrt{2 h r-h^2}+\sqrt{2 h r}\right]

and area

A=h \sqrt{2 h r-h^2}

, where

h

is the altitude from

A

B C

, then

\lim \limits_{h \rightarrow 0^{+}} \frac{A}{p^3}

View answer and explanation

Correct option: B

\frac{1}{128 r}

The correct answer is B (1/128r) because by simplifying the expression and evaluating the limit as h approaches 0, the result is 1/(128r).

180 Medium

180 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{\pi}{4 x}-\frac{\pi}{2 x\left(e^{\pi x}+1\right)}\right)

equals

View answer and explanation

Correct option: D

\frac{\pi^2}{8}

Option D (

\frac{\pi^2}{8}

) is correct because after simplifying the expression and applying L'Hôpital's rule, the limit evaluates to

\frac{\pi^2}{8}

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