I have an alternative solution without the use of the L'Hospital rule. Start as Paul suggested, but when in the form of
$ \lim_{x \to \infty} x \log \left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right) $
you can use the fact that
$ \lim_{y \to 1} \frac{\log y}{y - 1} = 1. $
Using this limit, the limit arithmetic and a limit of a composed function. All that helps you transform the limit above into
$ \lim_{x \to \infty} x \left(\frac{1+\tan(1/x)}{1-\tan(1/x)} - 1\right) = \lim_{x \to \infty} x \left(\frac{2\tan(1/x)}{1-\tan(1/x)}\right) = \lim_{x \to \infty} 2 \cdot \frac{\tan{1/x}}{\frac 1x} $ Going from the second part to the third one required yet another arithmetic to get rid of the denominator - that is obviously one, because it is continuous. The last bit can be solved using yet another known limit $ \lim_{y \to 0} \frac{\tan y}{y} = 1 $
So we know the limit is two, we apply the exponential function and get the result $e^2$.
Hope this helps as well.
(Sorry for the typesetting mess [no eq numbers], I have yet to learn how to work with this system.)