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Evaluate $$\lim_{x\to e}\left(\ln x\right)^{\frac{1}{1-\ln x}}$$

My work so far:

  1. Took $\ln$ of both sides;
  2. Used l'Hospital's rule on RHS, which gives $ln y = -1$
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  • $\begingroup$ How does L'Hospital give $\ln y=-1$? Please explain further what you did. What do you mean when you say "took ln of both sides"? What "side"? There are sides in an equation, but you don't have an equation, do you? $\endgroup$ – 5xum Jul 20 '17 at 7:31
  • $\begingroup$ You are almost done! If $\log\lim=-1$ then $\lim=\dfrac{1}{e}$ $\endgroup$ – Raffaele Jul 20 '17 at 9:52
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Hint. Note that by letting $1+\frac{1}{t}=\ln(x)$, $$\lim_{x\to e}\left(\ln x\right)^{\frac{1}{1-\ln x}}=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{-t}.$$ Now recall one of the main properties of the number $e$.

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The idea of taking the logarithm is good and actually you almost finished your job. Let's compute $$ \lim_{x\to e}\ln\Bigl((\ln x)^{\tfrac{1}{1-\ln x}}\Bigr)= \lim_{x\to e}\frac{\ln\ln x}{1-\ln x} $$ Applying l'Hôpital (it can be done, because the limit is in the form $0/0$): $$ \lim_{x\to e}\frac{\dfrac{1}{x\ln x}}{-\dfrac{1}{x}}= \lim_{x\to e}-\frac{1}{\ln x}=-1 $$ Therefore the given limit is $e^{-1}$: if $\lim_{x\to c} \ln f(x)=l$, then $\lim_{x\to c} f(x)=e^l$.

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Since $f(x)=\frac{1}{x}$ is a continuous function at $x=e$, we obtain:

$$\lim_{x\rightarrow e}(\ln{x})^{\frac{1}{1-\ln{x}}}=\lim_{x\rightarrow e}\left((1+\ln{x}-1)^{\frac{1}{\ln{x}-1}}\right)^{-1}=\left(\lim_{x\rightarrow e}(1+\ln{x}-1)^{\frac{1}{\ln{x}-1}}\right)^{-1}=e^{-1}=\frac{1}{e}.$$

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Without L'Hospital, using Taylor series

$$A=\sqrt[1-\log (x)]{\log (x)}\implies \log(A)=\frac {\log(\log(x))}{1-\log (x)}$$ $$\log(x)=1+\frac{x-e}{e}-\frac{(x-e)^2}{2 e^2}+O\left((x-e)^3\right)$$ $$\log(\log(x))=\frac{x-e}{e}-\frac{(x-e)^2}{e^2}+O\left((x-e)^3\right)$$ $$\log(A)=\frac{\frac{x-e}{e}-\frac{(x-e)^2}{e^2}+O\left((x-e)^3\right) }{-\frac{x-e}{e}+\frac{(x-e)^2}{2 e^2}+O\left((x-e)^3\right) }=-1+\frac{x-e}{2 e}+O\left((x-e)^2\right)$$ $$A=e^{\log(A)}=\frac{1}{e}+\frac{x-e}{2 e^2}+O\left((x-e)^2\right)$$ which shows the limit and how it is approached.

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