#### Answer

The solution is
$$\lim_{x\to e}\frac{\ln x-1}{x-e}=\frac{1}{e}.$$

#### Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to e}\frac{\ln x-1}{x-e}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to e}\frac{(\ln x-1)'}{(x-e)'}=\lim_{x\to e}\frac{\frac{1}{x}}{1}=\lim_{x\to e}\frac{1}{x}=\frac{1}{e}.$$