Q41 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
Given a real valued function ' $f$ ' such that
$f(x)=\left\{\begin{array}{cc}\frac{\tan ^2\{x\}}{x^2-[x]^2} & \text { for } x\gt0 \\ 1 & \text { for } x=0 \\ \sqrt{\{x\} \cot \{x\}} & \text { for } x \lt 0\end{array}\right.$
then
- A$\mathrm{LHL}=1$
- B$\mathrm{RHL}=\sqrt{\cot 1}$
- C$\lim _{x \rightarrow 0} f(x)$ exist
- Correct answerD$\lim _{x \rightarrow 0} f(x)$ does not exists
Solution
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \sqrt{\{-h\} \cot \{-h\}}$
$=\lim _{h \rightarrow 0} \sqrt{(1-h) \cot (1-h)}=\sqrt{\cot 1}$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{\tan ^2\{h\}}{h^2-[h]^2}=\lim _{h \rightarrow 0} \frac{\tan ^2 h}{h^2}=1$
$\therefore \lim _{x \rightarrow 0} f(x)$ does not exist.
Hence, the correct answer is option: 'D'.