Evaluating a limit with variable in the exponent For
$$\lim_{x \to \infty} \left(1- \frac{2}{x}\right)^{\dfrac{x}{2}}$$ 
I have to use the L'Hospital"s rule, right? So I get:
$$\lim_{x \to \infty}\frac{x}{2} \log\left(1- \frac{2}{x}\right)$$
And what now? I need to take the derivative of the log, is it:
$\dfrac{1}{1-\dfrac{2}{x}}$  but since there is x, I need to use the chain rule  multiply by the derivative of $\dfrac{x}{2}$  ? 

We know, $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$
Put $n=-\frac x2, \left(\lim_{x\to\infty}\left(1-\frac 2x\right)^{-\frac x2}\right)^{-1}=e^{-1}$
