# 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:
$$\lim_{y\to+\infty}\exp\left[\log \left(1-\frac{1+\log y}{y}\right)-\frac{\log y}{y-1}\right]=\lim_{y\to\infty}\exp(\log (1-0)-0)=\boxed{1}.$$
• The change of variables was mostly for myself, because I have an easier time visualizing limits to $\infty$ instead of limits to $1$. Doing $\exp\circ\log$ will help you out any time you are evaluating the limit of a product. It is especially useful when the product already involves a $\log$! – pre-kidney Jan 20 '15 at 17:09
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 $$\lim_{t\to 0}\dfrac{t\ln{t}-t\ln{(t+1)}}{t}=\lim_{t\to0}\left(\ln{\dfrac{t}{t+1}}-\dfrac{t}{t+1}+1\right)=1$$