Can I apply L'Hôpital to $\lim_{x \to \infty} \frac{x+\ln x}{x-\ln x}$? Can I apply L'Hôpital to this limit: 
$$\lim_{x \to \infty} \frac{x+\ln x}{x-\ln x}?$$
I am not sure if I can because I learnt that I use L'Hôpital only if we have $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and here $x-\ln x$ is $\infty-\infty$ and x tends to infinity.
 A: $x -\ln x$ 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.
A: Maybe it is better to just write:
$$\lim_{x \to \infty} \frac{x+\log x}{x-\log x}=\lim_{x \to \infty} \frac{1+\frac{\log(x)}{x}}{1-\frac{\log(x)}{x}}$$
A: You are quite correct to observe that $\infty-\infty$ makes no sense.  But that's not how one goes about evaluating $\lim_{x\to\infty}(x-\ln x)$.  Among the correct ways is to note that, for $x\gt2$, 
$$x-\ln x=1+\int_1^x\left(1-{1\over t}\right)dt\gt\int_2^x\left(1-{1\over t}\right)dt\gt\int_2^x{1\over2}dt={x-2\over2}\to\infty$$
The first inequality here simply throws away some stuff that's clearly positive; the second inequality amounts to subtracting from $1$ the largest value that $1\over t$ takes on for $t\ge2$, namely $1\over2$.
Once you know that $\lim_{x\to\infty}(x+\ln x)=\lim_{x\to\infty}(x-\ln x)=\infty$, then it's OK to apply L'Hopital and get
$$\lim_{x\to\infty}{x+\ln x\over x-\ln x}=\lim_{x\to\infty}{1+{1\over x}\over1-{1\over x}}={1+0\over1-0}=1$$
A: 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.
A: 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?
A: If you have a form other than $\frac{\infty}{\infty}$ or $\frac{0}{0}$, then you cannot use L'Hôpital's rule directly. If you have an indeterminate form in either or both of the numerator or denominator, then you need to resolve these individually to determine whether the fraction is actually in one of the allowed forms. Thus, the answer you're looking for is no, you cannot just use L'Hôpital's rule from the start. However, after you resolve the top and bottom, you will see that it is in one of the forms and so you can use the rule once that has been determined.
A: Let us look at the limit of
$$\lambda(x):=\frac{f(x)+g(x)}{h(x)+i(x)}$$ where all four functions tend to $\pm\infty$ without canceling the denominator.
We can rewrite
$$\lambda(x)=\frac{f(x)}{h(x)}\frac{1+\dfrac{g(x)}{f(x)}}{1+\dfrac{i(x)}{h(x)}},$$
where the ratios are now $\frac\infty\infty$ undeterminate forms. If those ratios do have limits, we can find them by L'Hospital (provided the conditions are fulfilled), by evaluating
$$\lim_{x\to\infty}\frac{f'(x)}{h'(x)}\frac{1+\dfrac{g'(x)}{f'(x)}}{1+\dfrac{i'(x)}{h'(x)}}.$$
But if those limits exist and if the lower one $\ne-1$, the expression is also the limit of
$$\mu(x):=\frac{f'(x)+g'(x)}{h'(x)+i'(x)}.$$
This establishes a generalized L'Hospital rule.

Indeed
$$\lim_{x \to \infty} \frac{x+\ln x}{x-\ln x}=\lim_{x \to \infty} \frac{1+\dfrac1x}{1-\dfrac1x}=1.$$
