Question:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)= \begin{cases}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right) & , \quad x \neq 0 \\ \alpha & , \quad x=0\end{cases}$ If $f$ is continuous at $\mathrm{x}=0$, then $\alpha$ is equal to :
Correct Option: 1
Solution:
For continuity
$\lim _{x \rightarrow 0} \frac{x^{3}}{4 \sin ^{4} x}\left(\ell n\left(1+2 x e^{-2 x}\right)-2 \ell n\left(1-x e^{-x}\right)\right)$
$=\alpha$
$\lim _{x \rightarrow 0} \frac{1}{4 x}\left[2 x e^{-2 x}+2 x e^{-x}\right]=\alpha$
$=\frac{1}{4}(4)=\alpha=1$