Why is $\lim_{x \rightarrow e}\frac{\ln{(x)}-1}{x-e}$ equal to $\frac{1}{e}$ As it says. Why? The best I could achieve is:
$$\lim_{x \rightarrow e}\frac{\ln(x)-\ln (e)}{x-e}$$
And the answer says it is equal to 
$$\frac{\text{d}}{\text{d}y}(\ln(x))$$
When $x$ is $e$ so it should be
$$\frac 1 x$$
And when $x$ is replaced by $e$ it is
$$\frac 1 e$$
Why is that? What converted the limit to a derivative?
Also no l'hopital.

The derivative of a real valued function $f$ at a point $y$ is given by $\lim_{x \to y} \frac{f(x) - f(y)}{x-y} = f'(y)$. In your case, if we have $f(x) = \ln x$ then we know $f'(x) = \frac{1}{x}$ and $f'(e) = \frac{1}{e}$. 
But $\displaystyle f'(e) = \lim_{x \to e} \frac{\ln x - \ln e}{x-e} = \lim_{x\to e} \frac{\ln x -1 }{x-e}$.
