How to find the limit $\lim_{x \rightarrow 1^+}\left (1 - \frac{1}{x}\right)^x \left( \log\left(1 - \frac{1}{x}\right) + \frac{1}{x - 1}\right)$ How can I find the limit $\lim_{x \rightarrow 1^+} \left (1 - \frac{1}{x}\right)^x \left( \log\left(1 - \frac{1}{x}\right) + \frac{1}{x - 1}\right)$?

I tried turning into a fraction so that I could apply L'Hopital's rule:
$\left(1 - \frac{1}{x}\right)^x \left(\frac{\log\left(1 - \frac{1}{x}\right)(x-1) + 1}{x - 1}\right)$

But that didn't seem to get me anywhere. Thanks.
Change variables to $y=\frac{x}{x-1}$, then $x=\frac{y}{y-1}$. Then your limit becomes:
$$
\lim_{y\to+\infty}\exp\left[\log (y-1-\log y)-\frac{y}{y-1}\log y\right]
$$
Now subtract $\log y$ from each of the terms to obtain:

$$
\lim_{y\to+\infty}\exp\left[\log \left(1-\frac{1+\log y}{y}\right)-\frac{\log y}{y-1}\right]=\lim_{y\to\infty}\exp(\log (1-0)-0)=\boxed{1}.
$$
let
$$I=\lim_{x\to 1^{+}}\left(1-\dfrac{1}{x}\right)^x\left(\ln{\left(1-\dfrac{1}{x}\right)}+\dfrac{1}{x-1}\right)$$
Let $x=t+1$$$\Longrightarrow I=\lim_{t\to 0}\left(1-\dfrac{1}{t+1}\right)^{t+1}\left(\ln{\dfrac{t}{t+1}}+\dfrac{1}{t}\right)=I_{1}\cdot I_{2}$$
since
$$I_{1}=\lim_{t\to 0^{+}}\left(1-\dfrac{1}{t+1}\right)^{t+1}=1$$
because  apply L'Hopital's 
$$I_{1}=\lim_{t\to 0^{+}}e^{(t+1)[\ln{t}-\ln{(t+1)}]}=1$$
and
$$I_{2}=\lim_{t\to 0}\left(\ln{\dfrac{t}{t+1}}+\dfrac{1}{t}\right)=1$$
becasue  apply L'Hopital's 
$$\lim_{t\to 0}\dfrac{t\ln{t}-t\ln{(t+1)}}{t}=\lim_{t\to0}\left(\ln{\dfrac{t}{t+1}}-\dfrac{t}{t+1}+1\right)=1$$