## Proving a different version of L'Hopital's Rule

Suppose $\displaystyle f: (a,b) \rightarrow \mathbb{R}$, $\displaystyle g: (a,b) \rightarrow \mathbb{R}$ are differentiable and that g(x) and g'(x) are always non-zero. Suppose that $\displaystyle lim_{x \rightarrow b^-}f(x)=0=lim_{x \rightarrow b^-}g(x)$ and that
$\displaystyle lim_{x \rightarrow b^-}\frac{f'(x)}{g'(x)}=\infty$ {Equation *}

then $\displaystyle lim_{x \rightarrow b^-}\frac{f(x)}{g(x)}=\infty$.

The usual version of L'Hopitals that I know how to prove involves us allowing to say (from continuous limits) that f(b)=g(b)=0 and thus we can then see that $\displaystyle lim_{x\rightarrow b} \frac{f(x)}{g(x)} = lim_{x\rightarrow b} \frac{f(x)-f(b)}{g(x)-g(b)} = lim_{x\rightarrow b} \frac{\frac{f(x)-f(b)}{x-b}}{\frac{g(x)-g(b)}{x-b}}=lim_{x\rightarrow b} \frac{f'(x)}{g'(x)}$

But here, g(x) is non-zero for all x, so I can't really do this. Furthermore, the fact that we're only dealing with a left limit makes things harder for me. I'm not quite sure where to start, I'm sure it's a simple solution but I can't see it.

I know the definition of {Equation *} is $\displaystyle \forall M >0$ $\displaystyle \exists \space \delta > 0 : 0<b-x<\delta \Rightarrow \frac{f'(x)}{g'(x)} > M$

I literally don't even know where to start anymore, I tried going straight from this definition and doing a bit of jiggery pokery but ran into a lot of problems. I've also taken a look at Cauchy's Mean Value Theorem but don't know what I can do with that, I can't say for certain that these functions are continuous on [a,b].

Can anyone help me find the path to a proof? Cheers.