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.

$$\lim_{x\to e} \frac{e-x}{\ln x-1} = -\lim_{x\to e}\frac{1}{\frac{\ln x-1}{x-e}} = -\lim_{x\to e}\frac{1}{\ln(\frac{x}{e})^{\frac{1}{x-e}}} = -\lim_{x\to e}\frac{1}{\ln{\bigg({\big(1+\frac{x-e}{e}\big)}^{\frac{e}{x-e}}\bigg)^{\frac{x-e}{e}\cdot\frac{1}{x-e}}}} = -\frac{1}{\ln e^{\frac{1}{e}}} = -e$$
With the well-known $$\lim_{f(x)\to 0}(1+f(x))^\frac{1}{f(x)} = e$$
