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 · Mixed difficulty

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211 Easy

211 of 930

Application +4 | -1

\alpha

is a repeated root of

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

, then

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

View answer and explanation

Correct option: B

a

Option B (

a

) is correct because the limit simplifies to

a

after substituting

a{x^2} + bx + c = a{\left( {x - \alpha } \right)^2}

and applying the limit.

212 Easy

212 of 930

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 3}} \frac{[x]^2-9}{x^2-9}

is equal to

View answer and explanation

Correct option: C

Does not exist

Option C (Does not exist) is correct because the left-hand limit and right-hand limit are not equal.

213 Easy

213 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{{5^x} - {5^{ - x}}}}{{2x}}

View answer and explanation

Correct option: A

\log 5

Option A (

\log 5

) is correct because the limit can be evaluated using L'Hopital's rule or by recognizing the derivative of an exponential function.

214 Medium

214 of 930

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.

215 Easy

215 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^2 x \cot ^2(2 x)}

is equ

View answer and explanation

Correct option: D

Option D (1) is correct because after rewriting the expression using trigonometric identities and simplifying, the limit evaluates to 1.

216 Medium

216 of 930

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}

217 Easy

217 of 930

Analyzing +4 | -1

If the normal to the curve

y=f(x)

x=0

be given by the equation

3 x-y+3=0

, then the value of

\lim \limits_{{x \rightarrow 0}} x^2\left\{f\left(x^2\right)-5 f\left(4 x^2\right)+4 f\left(7 x^2\right)\right\}^{-1}

, is

View answer and explanation

Correct option: B

-\frac{1}{3}

Option D (4, -4) is correct because it satisfies the given limit condition after applying L'Hopital's rule and using the information from the normal to the curve.

218 Medium

218 of 930

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.

219 Easy

219 of 930

Application +4 | -1

Let

f

be a non-zero continuous function satisfying

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

for all

x, y \in \mathbf{R}

. If

f(2)=9

then

f(3)

View answer and explanation

Correct option: B

Option B (27) is correct because the function f(x) is an exponential function, and given f(2)=9, we can determine that f(x) = 3^x, thus f(3) = 27.

220 Medium

220 of 930

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.

221 Easy

221 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: B

e

Option B (

e

) is correct because the limit can be rewritten in the form of the definition of

e

as n approaches infinity.

222 Easy

222 of 930

Analyzing +4 | -1

For

n \in N

, let

f_n(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)

(1+\sec 4 x) \ldots\left(1+\sec 2^n x\right)

. Then,

\lim \limits_{{x \rightarrow 0}} \frac{f_n(x)}{2 x}

View answer and explanation

Correct option: C

2^{n-1}

Option C,

2^{n-1}

, is correct because by repeatedly applying the identity

1 + \sec x = \frac{1 + \cos x}{\cos x} = \frac{2 \cos^2(x/2)}{\cos x}

, the function

f_n(x)

simplifies to

\tan(2^n x)

, and the limit can be evaluated using L'Hopital's rule or the standard limit

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

223 Medium

223 of 930

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

224 Medium

224 of 930

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.

225 Medium

225 of 930

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.

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226 Easy

226 of 930

Analyzing +4 | -1

\lim _{x \rightarrow 0}\left(\frac{3 x^2+2}{7 x^2+2}\right)^{1 / x^2}

is equal to:

View answer and explanation

Correct option: B

\frac{1}{e^2}

Option B (

\frac{1}{e^2}

) is correct because after simplifying the expression and evaluating the limit, the result is

\frac{1}{e^2}

227 Medium

227 of 930

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

228 Easy

228 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}}\left(2^{1 / 2} \cdot 2^{1 / 4} \cdot 2^{1 / 8} \ldots 2^{1 / 2^n}\right)

equals-

View answer and explanation

Correct option: B

2^1

Option B (

2^1

) is correct because the limit of the given expression simplifies to 2 raised to the power of 1, based on the sum of an infinite geometric series.

229 Medium

229 of 930

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.

230 Easy

230 of 930

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1+2^4+3^4+\ldots+n^4}{n^5}-\lim \limits_{{n \rightarrow \infty}} \frac{1^3+2^3+3^3+\ldots+n^3}{n^5}

, is

View answer and explanation

Correct option: A

\frac{1}{5}

Option A (

\frac{1}{5}

) is correct because the difference of the two limits is equal to

\frac{1}{5}

231 Easy

231 of 930

Analyzing +4 | -1

The values of constants a and

b

so that

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

, are

View answer and explanation

Correct option: A

a=1, b=-\frac{3}{2}

Option A (

a=1, b=-\frac{3}{2}

) is correct because these values satisfy the given limit condition after simplifying the expression.

232 Medium

232 of 930

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}

233 Medium

233 of 930

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}

234 Medium

234 of 930

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.

235 Hard

235 of 930

Analyzing +4 | -1

f(x)=\frac{e^{[x]+\mid x \mid}-3}{[x]+|x|+1}

, then: (where [.] represents greatest integer function)

View answer and explanation

Correct option: A

\lim \limits_{x \rightarrow 0^{+}} f(x)=-2

Option A is correct because the right-hand limit exists and equals -2, while the left-hand limit does not exist.

236 Easy

236 of 930

Understanding +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\left| x \right|}}{x}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand limit and the right-hand limit of |x|/x as x approaches 0 are not equal, hence the limit does not exist.

237 Easy

237 of 930

Application +4 | -1

f(x)= \begin{cases}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { for }-1 \leq x<0 \\ 2 x^2+3 x-2, & \text { for } 0 \leq x \leq 1\end{cases}

is continuous at

x=0

, then

k

is equal to

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because for the function to be continuous at x=0, the left-hand limit (LHL) must equal the right-hand limit (RHL).

238 Easy

238 of 930

Application +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{\int \nolimits_0^{{x^2}} \cos {t^2}dt}}{{x\sin x}}

View answer and explanation

Correct option: B

Option B (1) is correct because L'Hopital's rule can be applied twice to evaluate the limit, resulting in 1.

239 Easy

239 of 930

Application +4 | -1

f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7

, then

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

View answer and explanation

Correct option: B

\frac{53}{3}

Option B (53/3) is correct because it is obtained by applying L'Hopital's rule and using the given function and its derivative.

240 Easy

240 of 930

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because by applying L'Hopital's rule multiple times, the limit simplifies to

\frac{1}{2}

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