Can I apply L'Hôpital to $\lim_{x \to \infty} \frac{x+\ln x}{x-\ln x}$?

As an alternative to Stefano's calculation, note that the derivative of $x-\log x$ is $1-\frac1x$, which is $\ge \frac12$ whenever $x\ge 2$. Thus, by the mean value theorem we have $$ x-\log x \ge 2-\log 2 + \frac{x-2}{2} $$ for all $x\ge 2$, and the right-hand side of this clearly goes to $\infty$.

So $x-\log x\to \infty$ when $x\to\infty$.

It is also abundantly clear that $x+\log x$ goes to $\infty$ for $x\to\infty$, so you're allowed to try using L'Hospital on your fraction.

HINT Not sure you really need L'Hospital's Rule here. Note that you can divide the top and bottom of the fraction by $x$, to get $$ \frac{x+\ln x}{x-\ln x} = \frac{1+\frac{\ln x}{x}}{1-\frac{\ln x}{x}} $$ Does this make things easier?

$x -\ln x$ goes to $+\infty$ if and only if $e^{x-\ln x}$ does. And this is the case, since

$$e^{x-\ln x} = \frac{e^x}{x} \ge \frac{1+x+\dfrac{x^2}{2}}{x} \to +\infty $$

as $x \to +\infty$. So yes, you can apply de l'Hopital from the beginning.