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I've already seen other questions similar to this (for example, this one: How to prove l'Hospital's rule for $\infty/\infty$).

I'm trying to do the proof that, if $\lim \limits_{x \to \infty} f(x) = \lim \limits_{x \to \infty} g(x) = \infty$ and $\lim \limits_{x \to \infty} \frac {f'(x)}{g'(x)} = L$, then $$\lim \limits_{x \to \infty} \frac {f(x)}{g(x)} = L$$

I did the demonstration in a very similar way to the other question linked, using just the definition $\lim \limits_{x \to \infty}$ and the Cauchy's mean value theorem, until I got that:

For all $\epsilon>0$, there exists a $a>0$ such that:

$$ x>a \implies \left | \frac {f(x)-f(a)}{g(x)-g(a)} - L \right | < \epsilon$$

I don't know what sort of manipulation I could use to reach

$$ x>a \implies \left | \frac {f(x)}{g(x)} - L \right | < \epsilon '$$

(where $\epsilon'$ is not necessarily the same as $\epsilon$ in the first expression), which would complete the proof. I've seen the arguments used in this step in the other questions (like the one I linked) but, to be honest, I didn't understand them.

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For any $\epsilon > 0 $ there exists $a > 0$ such that if $x > a$ then

$$\left|\frac{f'(x)}{g'(x)}-L \right|< C\epsilon,$$

with $C = [2(1+|L|)]^{-1}$.

By the MVT there exists $c > a$ such that

$$\frac{f(x)}{g(x)}h(x):=\frac{f(x)}{g(x)}\frac{1- \frac{f(a)}{f(x)}}{1-\frac{g(a)}{g(x)}}=\frac{f(x) - f(a)}{g(x)-g(a)}= \frac{f'(c)}{g'(c)}.$$

Since $c > a$ 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 $\lim_{x \to \infty}h(x) = 1$. Hence, there exists $b > 0$ such that for $x > b$ we have $|h(x) - 1| < C\epsilon$ and $|h(x)| > 1/2.$

Whence, if $x > \max(a,b)$ then

$$\left|\frac{f(x)}{g(x)}-L\right| < 2(1 + |L|)C\epsilon = \epsilon.$$

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