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 $\dfrac{x}{2}$  ? 

If you insist in working with the l'Hospital:
$$\lim_{x\to\infty}\frac x2\log\left(1-\frac2x\right)=\lim_{x\to\infty}\frac{\log\left(1-\frac2x\right)}{\frac2x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac2{x^2}\frac1{1-\frac2x}}{-\frac2{x^2}}=-\lim_{x\to\infty}\frac1{1-\frac2x}=-1$$
Thus, the limit is $\,e^{-1}\,$ ...
