Question

The value of $e[\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}+\underset{x\to 1}{lim}{x}^{\frac{1}{1-x}}]$

Answer 1

Let say $l_1=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}$
$=\underset{x\to 0}{lim}{\left[{(1+(\frac{\sin x}{x}-1))}^{\frac{1}{\sin x/x-1}}\right]}^{-\frac{\sin x}{x}}=e^{-1}$
and $l_2=\underset{x\to 1}{lim}{x}^{\frac{1}{1-x}}=\underset{x\to 1}{lim}[1+(x-1){]}^{\frac{1}{1-x}}=e^{-1}$
So $l_1+l_2=2e^{-1}$
Hence $e\cdot (l_1+l_2)=e\left(2e^{-1}\right)=2$