Evaluate the limit $\lim _{x\to e}\left(\frac{1-\log _e\left(x\right)}{x-e}\right)$ without l'Hôpital's Rule 
$\lim _{x\to e}\left(\frac{1-\log _e\left(x\right)}{x-e}\right)$

I have tried to use this fact: $\lim _{x\to 0}\left(\frac{\left(a^x-1\right)}{a}\right)\:=\:\ln a$.
 A: Hint Identify the expression in the limit as the difference quotient of a particular function $f$ at a particular point $x_0$, and hence as the value $f'(x_0)$.
A: $$\lim_{x\to e}\dfrac{1-\ln x}{x-e}=-\dfrac1e\cdot\lim_{x\to e}\dfrac{\ln\left(1+\dfrac{x-e}e\right)}{\dfrac{x-e}e}=?$$ as $1=\ln e$
A: Let $f(x):= \log_e(x)$. Then:
$\frac{1-\log _e(x)}{x-e}= -\frac{f(x)-f(e)}{x-e} \to -f'(e)$.
We have not used Lopitals rule, we only used the definition of the derivative.
A: $$
\begin{align}
\lim _{x\to e}\frac{1-\log _e(x)}{x-e}&= -\lim _{x\to e}\frac{\log_{e}(x)-1 }{x-e} \\
&= -\frac{d}{dx}\Bigg\vert_{x=e} \log_e(x) \\
&= -\frac{1}{e}
\end{align}
$$
A: I'll try a different approach actually not related to L'Hôpital rule.
Suppose the limit $L$ exists, then calculate $e^L$
$$e^L=\lim _{x\to e}\exp\left(\frac{1-\ln x}{x-e}\right)=$$
$$=\lim _{x\to e}\exp(1-\ln x)^{1/(x-e)}=\lim _{x\to e}\left(\dfrac{e}{x}\right)^{1/(x-e)}=$$
$$=\lim _{x\to e}\left(1+\dfrac{e}{x}-1\right)^{1/(x-e)}=\lim _{x\to e}\left(1+\dfrac{e-x}{x}\right)^{(-1/x)(x/(e-x))}=$$
$$=\lim _{x\to e}\left[\left(1+\dfrac{e-x}{x}\right)^{x/(e-x)}\right]^{-1/x}=$$
$$=e^{-1/e}$$
Then $L=\ln e^L=\ln e^{-1/e}=-1/e$
A: With the change of variable $x=e^{t+1}$,
$$\lim_{x\to e}\frac{1-\ln x}{x-e}=\lim_{t\to0}\frac{-t}{e^{t+1}-e}=-\frac1e\lim_{t\to0}\frac{t}{e^{t}-1}=-\frac1e\frac1{\lim_{t\to0}\dfrac{e^{t}-1}t}.$$
Then use the given fact with $a=e$.
