What is the limit as x approaches a for, (ln(x)-ln(a))/(x-a)
As Hollywood posted, they want you to see that it's the derivative of f(x) = ln(x) at x = a, and so is 1/a. I've usually seen that done by defining the natural log as an integral, and then taking its derivative using the Fundamental Theorem of Calculus.
However, if you only know about the derivative of $\displaystyle e^x$, and have $\displaystyle ln(x)$ defined as its inverse, you could do this:
Let $\displaystyle f(x) = e^x$. Since $\displaystyle f^{-1}(f(x)) = x \ \forall x \in \mathbb{R}$, take derivatives to get $\displaystyle (f^{-1})'(f(x))(f'(x)) = 1 \ \forall x \in \mathbb{R}$.
Since $\displaystyle f(x) = e^x$, that reads $\displaystyle (f^{-1})'(f(x)) = 1/f(x) \ \forall x \in \mathbb{R}$, so
$\displaystyle (f^{-1})'(e^x) = 1/(e^x) \ \forall x \in \mathbb{R}$, so, since the image of f is all positive reals, have
$\displaystyle (f^{-1})'(t) = 1/t \ \forall t> 0, t \in \mathbb{R}$.
Also, could do this:
$\displaystyle \lim_{x \to a} \frac{ln(x)-ln(a)}{x-a} = \lim_{x \to a} \frac{ln(x/a)}{a\left(\frac{x}{a}-1\right)} $
$\displaystyle = \frac{1}{a} \ \lim_{x \to a} \frac{ln(x/a)}{\frac{x}{a}-1} = \frac{1}{a} \ \lim_{t \to 1} \frac{ln(t)}{t-1}$
$\displaystyle =$(substituting $\displaystyle t = e^s$) $\displaystyle \frac{1}{a} \ \lim_{s \to 0} \frac{ln(e^s)}{e^s-1} = \frac{1}{a} \ \lim_{s \to 0} \frac{s}{e^s-1}$
$\displaystyle = \frac{1}{a} \ \lim_{s \to 0} \frac{1}{\frac{e^s-1}{s}} = \frac{1}{a} \ \frac{1}{\lim_{s \to 0} \frac{e^s-1}{s}}$
$\displaystyle = \frac{1}{a} \ \frac{1}{1} = \frac{1}{a}$