Calculate $\lim\limits_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ How can we find $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ without using:
(a) L'Hôpital's rule, (b) the limit $\displaystyle \lim_{h \to 0}\frac{e^h-1}{h} = 1$, and (c) the fact that $\displaystyle \frac{d}{dx}\left(e^x\right) = e^x$. 
The reason for the conditions is that with this limit I'm trying to prove (c), and I've done so with (b) and I gather it would be circular to use (a). So that's that. Also, I would appreciate if you could share one or more ways of proving that the derivative of $e^x$ is $e^x$. Thanks a lot for your time.  

If $e^x=\sum_{n\geq0}x^n/n!$, then you can show easily that $e^{x+y}=e^xe^y$. Then $$\frac{e^{x+h}-e^x}{h} = \frac{e^h-1}{h}e^x,$$ and to conclude that $\frac{d}{dx}e^x=e^x$ we need only then show that $$\frac{e^h-1}{h}\to1$$ if $h\to0$. Using your definition, that is easy.
