calculus - Can someone show how to find the limits of these functions using L'Hopitals rule?

I understand how to use L'Hopitals rule for the most part but these two problems confuse me to no end. I would appreciate it if someone could show me how they are to be done.

first one...

$$\lim_{x\to\infty}xe^{-x}$$

second one...

$$\lim_{x\to\frac{\pi}{2}^-}\frac{\tan x}{\ln(\frac{x}{2} - x)}$$

Answer

Gerry Myerson’s comment should take care of the first problem. The second one has to be misstated: I’ve very little doubt that it should be

$$\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}\;.$$

Corrected: If so, apply l’Hospital’s rule once and do a little simplification:

$$\begin{align*}
\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\tan x}{\ln\left(\frac{\pi}2-x\right)}&=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{\sec^2 x}{\frac{-1}{\frac{\pi}2-x}}\\\\
&=\lim_{x\to\left(\frac{\pi}2\right)^-}\sec^2 x\left(x-\frac{\pi}2\right)\\\\
&=\lim_{x\to\left(\frac{\pi}2\right)^-}\frac{x-\frac{\pi}2}{\cos^2 x}\;.
\end{align*}$$

Now apply l’Hospital’s rule one more time.