How do I evaluate the limiting value of $\lim_\limits{x\to e}\left(\ln x\right)^{\frac{1}{1-\ln x}}$

Question

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 $ln y = -1$

Answer 1

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 $e$.

Answer 2

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.

Answer 3

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 $\lim_{x\to c} f(x)=e^l$.

Answer 4

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

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