Question about limits and Mean Value Theorem 
Let $f:(a,b) \rightarrow \mathbb{R}$ and  $g:(a,b) \rightarrow \mathbb{R}$ be differentiable on (a,b) with $g'(x) \neq 0$ for all $x$ in $(a,b)$. Suppose  $\lim_{x \to b-}\dfrac{f'(x)}{g'(x)}$ exists.
  Show that for $\epsilon >0$ with $b-\epsilon >a$ we have $g(b- \epsilon) \neq g(b)$ and that $$ \lim_{\epsilon \to 0+}\dfrac{f(b)-f(b-\epsilon)}{g(b)-g(b-\epsilon)}= \lim_{x \to b-}\dfrac{f'(x)}{g'(x)}$$ carefully quoting the results you use:  

My Attempt
Define $$h(x)=f(x)(g(b)-g(b-\epsilon))-g(x)(f(b)-f(b-\epsilon))$$ Now $$h(b)=f(b)g(b)-f(b)g(b-\epsilon)-g(b)f(b)+g(b)f(b-\epsilon)=g(b)f(b-\epsilon)-f(b)g(b-\epsilon)$$ And $$h(\b- \epsilon)=f(b- \epsilon)g(b)-f(b- \epsilon)g(b- \epsilon)-g(b- \epsilon)f(b)+g(b- \epsilon)(f(b- \epsilon)=f(b- \epsilon)g(b)-g(b- \epsilon)f(b)$$
So Rolles Theorem cannot be applied....
How would I approach this problem??
Any help would be much appreciated.
 A: If we assume that $f(x), g(x)$ are continuous on $b$ and set $$F(x) = f(b) - f(b - x), G(x) = g(b) - g(b - x)$$ then we see that the functions $F, G$ are defined and continuous on $[0, b - a)$ and also differentiable on $(0, b - a)$. Since $g'(x) \neq 0$ in $(a, b)$ it follows that $G(x)\neq 0$ for $x \in (0, b - a)$. Again we have $$F'(x) = f'(b - x), G'(x) = g'(b - x)$$ and we also see therefore that $G'(x) \neq 0$ for all $x \in (0, b - a)$. Since the limit $$\lim_{x \to b^{-}}\frac{f'(x)}{g'(x)} = L$$ exists we can say that $$\lim_{x \to 0^{+}}\frac{F'(x)}{G'(x)}$$ exists and is equal to $L$. Since $F(x), G(x)$ both tend to $0$ as $x \to 0^{+}$ we are in a perfect situation to apply L'Hospital's rule and get $$\lim_{x \to 0^{+}}\frac{F(x)}{G(x)} = \lim_{x \to 0^{+}}\frac{F'(x)}{G'(x)} = L$$ or $$\lim_{x \to 0^{+}}\frac{f(b) - f(b - x)}{g(b) - g(b - x)} = \lim_{x \to b^{-}}\frac{f'(x)}{g'(x)}$$ Now you can see that your question is a statement of L'Hospital's rule in disguise. A proof is available in my blog post. This proof uses Cauchy's Mean Value Theorem.
