Is there a proof for L'Hôpital's Rule for limits approaching infinity? L'Hopital's Rule states that:
For two differentiable functions $f$ and $g$, where $g'(x)\neq 0$, such that
$$\lim_{x\to a} f(x)=0$$
$$\lim_{x\to a} g(x)=0$$
We can say that:
$$\lim_{x\to a} {f(x)\over g(x)}= \lim_{x\to a} {f'(x)\over g'(x)}\\ $$
NOTE: If you already know the proof and don't want to read all this, skip all the way down to $\blacksquare_{1.1}$
PROOF 1.1:
Let $f$ and $g$ be continuous functions on $[a,b]$ and differentiable on $(a,b)$. 
Also, assume $g'(x)\neq 0$ on $(a,b)$ and $g(b)\neq g(a).\\$
Proposition 1.1.1: There exists some point $c$ within the open interval $(a,b)$ such that:
${f'(c)\over g'(c)} = {{f(b) - f(a)}\over {g(b) - g(a)}}$
Proof 1.1.1 $\quad \triangleright$
Let $$\space h(x)= f(x)-f(a) - {{f(b) - f(a)}\over {g(b) - g(a)}}\cdot (g(x)-g (a))\\$$
By simply substituting values we can clearly see that $h(a)=h(b)=0$. 
Now because $f(a)$, $f(b)$, $g(a)$ and $g(b)$ are constants, we can therefore say that much like $f$ and $g$,  $h$ is also continuous on $[a,b]$ and differentiable on $(a,b).\\$
If we differentiate  $h$ w.r.t. $x$, we get the following:
$$h'(x) = f'(x) - g'(x)\cdot {{f(b) - f(a)}\over {g(b) - g(a)}}\\$$
Using Rolle's Theorem (which I porpusely won't prove as the question is long enough as it is) we can say that there exists a $c$ in $(a,b)$ such that $h'(c)=0$
Thus we can say that
$0 = f'(c) - g'(c)\cdot {{f(b) - f(a)}\over {g(b) - g(a)}} $
Hence showing us that
${f'(c)\over g'(c)} = {{f(b) - f(a)}\over {g(b) - g(a)}}$
$$\blacksquare_{1.1.1}$$
$\triangleleft \\$
Remember that $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)=0$. 
Where $a$ is finite. 
We also said $g(x)\neq 0$.
Therefore we'll Let
$L:= \lim_{x\to a} {f'(x)\over g'(x)} \\$
We're also going to define the functions $F$ and $G$. 
$F(x) = f(x) \Longrightarrow x\neq a$
$F(x) = 0 \Longrightarrow x = a$
Similarly
$G(x) = g(x) \Longrightarrow x\neq a$
$G(x) = 0 \Longrightarrow x = a$
Because $F$ and $G$ are defined at $x=a$, they are continuous at $a$. (Unlike $f$ and $g$)
This means that for $x>a$, the functions  $F$ and $G$ are differentiable on the open interval  $(a,x)$ and continuous on the closed interval $[a,x]. \\$
Using what we showed in Proof 1.1.1,  we can state the following equality to be true:
$${F'(c)\over G'(c)} = {{F(x) - F(a)}\over {G(x) - G(a)}}$$
Due to the fact that $F(a)=0$ and $G(a)=0$, we can thus say
$${F'(c)\over G'(c)} = {{F(x)}\over {G(x)}}$$ 
Now since $c$ is within the interval $(a,x)$, we can say $a<c<x$. 
This means that
$x\rightarrow {a^+} \Longrightarrow c\rightarrow {a^+}$
Therefore because $F$ and $G$ are simply the functions $f$ and $g$ respectively where $x=a$ is defined. It's correct for us to then say
$$\lim_{x\to a^+} {f(x)\over g(x)} = \lim_{x\to a^+} {F(x)\over G(x)} = \lim_{c\to a^+} {F'(c)\over G'(c)} $$
Also
$$\lim_{c\to a^+} {F'(c)\over G'(c)} = \lim_{c\to a^+} {f'(c)\over g'(c)}\\$$
If we notice... 
$$L:=\lim_{c\to a^+} {f'(c)\over g'(c)}= \lim_{x\to a^+} {f'(x)\over g'(x)}$$
Hence showing us that
$$\lim_{x\to a} {f(x)\over g(x)}= \lim_{x\to a} {f'(x)\over g'(x)} $$
$$\blacksquare_{1.1}\\$$
Isn't this only true when $\lim_{x\to a} {f(x)\over g(x)}={0\over 0}$? 
How would you prove $\lim_{x\to a} {f(x)\over g(x)}= \lim_{x\to a} {f'(x)\over g'(x)} $ for limits such as:
$\lim_{x\to a} {f(x)\over g(x)}={\pm\infty\over \pm\infty}$?
 A: Suppose $f(x),g(x) \to +\infty$ as $x \to a+$ and 
$$ \displaystyle \lim_{x \to a+} \frac{f'(x)}{g'(x)} = L.$$
For any $\epsilon > 0 $ there exists $\delta_1 > 0$ such that if $a < x < a + \delta_1$ then
$$\left|\frac{f'(x)}{g'(x)}-L \right|< C\epsilon,$$
with $C = [2(1+|L|)]^{-1}$.
Fix $x_1 < a + \delta_1.$ By the MVT there exists $c$ such that
$$\frac{f(x)}{g(x)}h(x):=\frac{f(x)}{g(x)}\frac{1- \frac{f(x_1)}{f(x)}}{1-\frac{g(x_1)}{g(x)}}=\frac{f(x) - f(x_1)}{g(x)-g(x_1)}= \frac{f'(c)}{g'(c)}.$$
Since $x < c < x_1 < a + \delta_1 $ we have 
$$C\epsilon > \left|\frac{f'(c)}{g'(c)}-L \right| = \left|\frac{f(x)}{g(x)}h(x)-L \right|.$$
Hence, using the reverse triangle inequality
$$C\epsilon > \left|\frac{f(x)}{g(x)}h(x)-Lh(x) + L h(x)-L \right|\geqslant \left|\frac{f(x)}{g(x)}-L\right||h(x)| - |L| |h(x)-1|, $$
and
$$ \left|\frac{f(x)}{g(x)}-L\right||h(x)| < C\epsilon + |L| |h(x)-1|.$$
Note that $\lim_{x \to a+}h(x) = 1$. Hence, there exists $\delta_2 > 0$ such that for $a < x < a + \delta_2$ we have  $|h(x) - 1| < C\epsilon$ and $|h(x)| > 1/2.$
Whence, if $a < x < a +  \min(\delta_1,\delta_2)$ then
$$\left|\frac{f(x)}{g(x)}-L\right| < 2(1 + |L|)C\epsilon = \epsilon.$$
Therefore,
$$\lim_{x \to a+} \frac{f(x)}{g(x)} = \lim_{x \to a+} \frac{f'(x)}{g'(x)} .$$
