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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...

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

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    Yes, $1/(e^x)$, which has limit zero as $x\to\infty$.2012-11-29

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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.

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    @DoesTheLimExist: No problem; you’re welcome. (And yes, it’s a good idea to have the derivatives of the trig functions on tap.)2012-11-29