How to prove L'Hospital's rule using $\varepsilon$-$\delta$ method?

Question

For $\varepsilon$-$\delta$ proofs, basically we need to find a $\delta$ such that $|F(x)-L|<\epsilon$ whenever, $0<|x-a|<\delta$ (for a small positive number $\epsilon$).

To prove L'Hospital's rule (for when numerator and denominator function both tend to $0$ as $x\rightarrow a^{+}$) let us assume $F(x)=\frac{f(x)}{g(x)}$. Where, $\lim_{x\rightarrow a^{+}}f(x)=0$ and $\lim_{x\rightarrow a^{+}}g(x)=0$.

I claim that $L=\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}$. Now I need to prove this $L$ is indeed the limit.

$$\left|\frac{f(x)}{g(x)}-\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}\right|<\epsilon.$$

But after this I cannot understand how to find $\delta$ as a function of $\epsilon$, so that I can complete the proof. How should I proceed?

Do you know the Cauchy version of the mean value theorem? http://mathworld.wolfram.com/CauchysMean-ValueTheorem.html — Thomas, Aug 20 '17 at 08:17
Just note that $f(a)= g(a)=0$ and replace $f(x)/g(x)$ by the remaining expression. — Thomas, Aug 20 '17 at 08:19
@Thomas I get $$|\dfrac{f'(c)}{g'(c)}-\lim_{x \rightarrow a^{+}}\dfrac{f'(x)}{g'(x)}|<\epsilon$$ (where $c\in[a,x]$) — , Aug 20 '17 at 08:21
Yes. And by assumption the quotient of the derivatives converges (otherwise the limit would not make sense). What else do you need? — Thomas, Aug 20 '17 at 08:22

score 2 · Accepted Answer · edited Sep 30 '19 at 23:28

You need to assume that $$\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$$ exists. This means that for any $\varepsilon > 0 $ there is $\delta > 0 $ such that $$\tag{$\ast$}\quad\quad\quad\quad\left| \frac{f^\prime(y)}{g^\prime(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|< \varepsilon$$ whenever $|y-a|< \delta$.

Now the Cauchy version of the mean value theorem states that, if $f$ and $g$ are differentiable on $[b,a]$, say, and $x\in [b,a]$, and if $g(x)-g(a)\neq 0$, then there exists $c\in (x, a)$ such that $$\frac{f(x)-f(a)}{g(x)-g(a) }=\frac{f^\prime(c)}{g^\prime(c)}$$ I assume this to be known, the tricky part is to be able to choose $c$ for both the numerator and denominator simultaneously. This also is true only if $a\neq \infty$, so the following proof of l'Hopital only works under this additional assumption.

Now let's look at the claim. Choose $\varepsilon > 0$. You want to estimate $$\left| \frac{f(y)}{g(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right| < \varepsilon$$ for $y$ close to $a$. Choose $\delta >0 $ such that $(*) $ is true for your choice of $\varepsilon$. By the mean value theorem (Cauchy version) and the assumption $f(a)=g(a) =0$, the difference you are looking at is just the same as $$\left| \frac{f^\prime(c)}{g^\prime(c)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|$$ for some $c\in (y,a)$ which is $<\varepsilon$ by $(*)$ if only $|y-a| < \delta$.

How to prove L'Hospital's rule using $\varepsilon$-$\delta$ method?

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