Does L'Hopitals Rule Work for Evaluating Difference of Limits? Question : Given the function $f(x) = x^2 - e^x$, find the limit of $f$ as it approaches positive and negative infinity

Finding the limit of $f$ as it approaches $-\infty$ is simple and the answer your get is $$\lim_{x\to-\infty} f(x) = +\infty$$
Finding the limit as $f$ approaches $+\infty$ is not as straightforward. Evaluating by direct substitution you get an indeterminate form
$$\lim_{x\to+\infty} f(x) = (\infty)^2 - \infty = \infty - \infty$$
Now I realized L'Hoptial's Rule is the general go to method for evaluating indeterminate forms, but not usually for indeterminate forms of this type. Normally L'Hopital's Rule is used for indeterminate forms of quotients e.g ($\ \frac{0}{0}\  $ or  $\ \frac{\infty}{\infty}\  $), not differences e.g ($\infty - \infty$)
However when I tried to apply L'Hopitals Rule for the differences of limits (and not the quotient) as you will see below I still arrived at the right answer.
This is what I did : 
Let $g(x) = x^2$, $h(x) = e^x$ $\implies f(x) = g(x) - h(x)$
$$\lim_{x\to+\infty} f(x) = \lim_{x\to+\infty} g(x) - \lim_{x\to+\infty} h(x)$$
$$\lim_{x\to+\infty} g(x) - \lim_{x\to+\infty} h(x) = \infty - \infty$$

L''Hopital's Rule states that given :
$$(\lim_{x\to\infty} g(x) = \infty) \land (\lim_{x\to\infty} h(x) = \infty) \land (\lim_{x\to\infty} \frac{g'(x)}{h'(x)} = L) \implies \lim_{x\to\infty} \frac{g(x)}{h(x)} = L$$
What I've essentially done below is this (which I'm not sure is valid by L'Hopital's Rule):
$$(\lim_{x\to\infty} g(x) = \infty) \land (\lim_{x\to\infty} h(x) = \infty) \land (\lim_{x\to\infty} g'(x) - h'(x) = L) \implies \lim_{x\to\infty} g(x) - h(x) = L$$

Since L'Hopital's Rule works for the $n^{th}$ derivative, I took the second-derivative of $f(x)$ and evaluated the limits for the differences. Again I don't know if you can even do this with L'Hopital's Rule, but I tried it nonetheless.
$$\lim_{x\to+\infty} g''(x) - \lim_{x\to+\infty} h''(x) = \lim_{x\to+\infty} 2 - \lim_{x\to+\infty}e^x = -\infty$$
Therefore by a 'modified version' of L'Hopital's Rule (if you can call it that) I get : $$\lim_{x\to+\infty} f(x) = -\infty$$
Plotting this on WolframAlpha shows that this is indeed the case :
Wolfram|Alpha Plot
So is this just a spooky coincidence, or does L'Hopital's Rule work for determining the differences of limits?
 A: Yet the general answer is ‘No’. Here is a counter-example: 
Set $f(x)=\sqrt{x^2+x}-x$. Applying ‘L'Hospital's rule’ to this difference would give
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\biggl(\frac{2x+1}{2\sqrt{x^2+x}}-1\biggr)=0. $$
However, simply rationalising the numerator, you get
$$f(x)=\frac{x}{\sqrt{x^2+x}+x}\xrightarrow[x\to+\infty]{}\frac12.$$
A: If $a < b$ are positive real numbers and $g(x) = ax$, $h(x) = bx$, then $g$, $h \to \infty$ and $g' - h' \to a - b$, but $g - h \to -\infty$.
The problem (which your example skirts) is, $(g - h)' = g' - h'$. Consequently, if $g' - h'$ approaches a finite, non-zero limit, you're guaranteed that $g - h$ is unbounded.

In your case, one approach is to factor $e^{x}$:
$$
x^{2} - e^{x} = e^{x}(x^{2}e^{-x} - 1),
$$
and to use l'Hôpital to show the factor in parentheses goes to $-1$.
A: The answer is not (try to find a counterexample, there are quite many) and to apply L'Hospital's Rule to the function $f(x)=g(x)-h(x)$ when both $g$ and $h$ tend to infinity you can do the following
$$f(x)=g(x)-h(x)=\frac{1}{\frac{1}{g(x)}}-\frac{1}{\frac{1}{h(x)}}=\frac{\frac{1}{h(x)}-\frac{1}{g(x)}}{\frac{1}{g(x)h(x)}}$$
so you go to the form $\frac 00$
