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Limits: Let Cases 4pt 4pt Cases Step Function Fractional

JEE Maths question with a full step-by-step solution.

Question

Let

f(x)=\begin{cases}\dfrac{\tan^{2}\{x\}}{x^{2}-[x]^{2}}, & x>0\\[4pt] 1, & x=0\\[4pt] \sqrt{\{x\}\cot\{x\}}, & x<0\end{cases}

where

[x]

is the step up function and

\{x\}

is the fractional part function of

x

, then

\displaystyle\lim_{x\to 0^{+}}f(x)=1

correct

\displaystyle\lim_{x\to 0^{-}}f(x)=1

\cot^{-1}\left(\displaystyle\lim_{x\to 0^{-}}f(x)\right)^{2}=1

correct

DNone

Solution

Step 1: Compute the RHL (

x\to 0^{+}

). For small

x>0

\{x\}=x

and

[x]=0

, so

x^{2}-[x]^{2}=x^{2}

\lim_{x\to 0^{+}}\dfrac{\tan^{2}x}{x^{2}}=\lim_{x\to 0^{+}}\left(\dfrac{\tan x}{x}\right)^{2}=1

Option (1) is correct. Step 2: Compute the LHL (

x\to 0^{-}

). For

x

slightly less than

0

(between

-1

and

0

\{x\}=1+x

, so

\{x\}\to 1^{-}

. Let

h=-x\to 0^{+}

, so

\{x\}=1-h\to 1^{-}

\lim_{h\to 0^{+}}\sqrt{(1-h)\cot(1-h)}=\sqrt{1\cdot\cot 1}=\sqrt{\cot 1}

So LHL

=\sqrt{\cot 1}

, not

1

. Option (B) is wrong. Step 3: Check option (C):

\cot^{-1}\left(\sqrt{\cot 1}\right)^{2}=\cot^{-1}(\cot 1)=1

Option (3) is correct. Correct answers: (1) and (3)

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