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:

$$ \lim_{y\to+\infty}\exp\left[\log \left(1-\frac{1+\log y}{y}\right)-\frac{\log y}{y-1}\right]=\lim_{y\to\infty}\exp(\log (1-0)-0)=\boxed{1}. $$

  • $\begingroup$ To clarify: You can subtract log(y) from each of the terms because you're looking at the limit of the difference of two terms? And the last step is using L'Hopital's rule (or just prior knowledge) to say that y grows faster than log(y)? Also, is there a heuristic you use to know when to apply the change of variables or exp(log()) techniques? $\endgroup$ – user162988 Jan 19 '15 at 20:28
  • $\begingroup$ 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 $\log$! $\endgroup$ – 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 $$\lim_{t\to 0}\dfrac{t\ln{t}-t\ln{(t+1)}}{t}=\lim_{t\to0}\left(\ln{\dfrac{t}{t+1}}-\dfrac{t}{t+1}+1\right)=1$$


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