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Limits — JEE Maths practice question

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

Question
Let f(x)={tan⁡2{x}x2−[x]2,x>01,x=0{x}cot⁡{x},x<0f(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][x] is the step up function and {x}\{x\} is the fractional part function of xx, then
Alim⁡x→0+f(x)=1\displaystyle\lim_{x\to 0^{+}}f(x)=1correct
Blim⁡x→0−f(x)=1\displaystyle\lim_{x\to 0^{-}}f(x)=1
Ccot⁡−1(lim⁡x→0−f(x))2=1\cot^{-1}\left(\displaystyle\lim_{x\to 0^{-}}f(x)\right)^{2}=1correct
DNone
Solution
Step 1: Compute the RHL (x→0+x\to 0^{+}). For small x>0x>0: {x}=x\{x\}=x and [x]=0[x]=0, so x2−[x]2=x2x^{2}-[x]^{2}=x^{2}.
lim⁡x→0+tan⁡2xx2=lim⁡x→0+(tan⁡xx)2=1\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→0−x\to 0^{-}). For xx slightly less than 00 (between −1-1 and 00): {x}=1+x\{x\}=1+x, so {x}→1−\{x\}\to 1^{-}. Let h=−x→0+h=-x\to 0^{+}, so {x}=1−h→1−\{x\}=1-h\to 1^{-}.
lim⁡h→0+(1−h)cot⁡(1−h)=1⋅cot⁡1=cot⁡1\lim_{h\to 0^{+}}\sqrt{(1-h)\cot(1-h)}=\sqrt{1\cdot\cot 1}=\sqrt{\cot 1}
 So LHL =cot⁡1=\sqrt{\cot 1}, not 11. Option (B) is wrong. Step 3: Check option (C):
cot⁡−1(cot⁡1)2=cot⁡−1(cot⁡1)=1\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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