Solve $\lim_{x \rightarrow e} \frac{e-x}{\ln x -1}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = \frac{e-x}{\ln(x)-\ln(e)} = \frac{e-x}{\ln(\frac{x}{e})} = ???$$
What do I do next? I think I could use $\lim \frac{\log_a x}{x} = 0, a>1$, but I'm not sure how.

To avoid the derivative (since this is effectively L'Hospital's rule), notice that
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = -e\lim_{x \rightarrow e} \frac{\frac{x}{e} - 1}{\ln \frac{x}{e} } = -e \lim_{y \to 1} \frac{y-1}{\ln y}.$$
Using the well-known inequality $(y-1)/y \leqslant \ln y \leqslant y - 1$ it follows that
$$1 \leqslant \frac{y-1}{\ln y} \leqslant y,$$
and applying the squeeze theorem
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1}  = -e \lim_{y \to 1} \frac{y-1}{\ln y} = -e \cdot 1 = -e.$$
