Evaluate $\lim_{x\rightarrow 0}(1-1/x)^x$ using l'Hospital's rule. The problem is $$\lim_{x\rightarrow 0}(1-1/x)^x$$

I took the natural log and got:

$x\ln (1-(1/x))  $

What do i do now? My brain is clogged it seems I forgot how to simplify the expression inside $\ln$, help?
\begin{align}
\lim_{x \to 0} (1 - 1/x)^x &amp; =  \lim_{x \to 0} \exp(x\ln(1 - 1/x)\\
 &amp;  \\
 &amp; = \exp \left( \lim_{x \to 0} \frac{\ln(1 - 1/x)}{1/x} \right) \\ 
 &amp;  \\
 &amp; = \text{l'Hospitals rule} \\
 &amp;  \\
 &amp; = \exp \left(\lim_{x \to 0} -\frac{x}{x-1}\right) \\ 
 &amp;  \\
 &amp; =1
\end{align}
Write:
$$ lny = \frac{ln [1-(1/x)]}{\frac{1}{x}} $$

Now, if you are looking for x goes to infinity, you have type $0/0$, and you have  and if x goes to zero you have type $\infty/\infty $. In both cases, L'Hopital applies: take derivatives and evaluate.