$$L=\lim_{x\to e} (2-\ln x)^\frac{1}{x^2-e^2}$$
Here are my calculations:
\begin{align}
\lim_{x\to e} (2-\ln x)^\frac{1}{x^2-e^2}&=\lim_{x\to e} (1+1-\ln x)^\frac{1}{x^2-e^2}\\ &=\lim_{x\to e} ((1+1-\ln x)^\frac{1}{1-\ln x})^\frac{1-\ln x}{x^2-e^2}\\ &=e^{\lim_{x\to e}\frac{1-\ln x}{x^2-e^2}}.
\end{align}
Then, for the exponent,
$$
\lim_{x\to e} \frac{1-\ln x}{x^2-e^2}=\lim_{x\to e} \frac{\ln e-\ln x}{e^{2\ln x}-e^{2\ln e}}=\lim_{x\to e} \frac{\ln e-\ln x}{e^{2\ln e}(e^{2\ln x-2\ln e}-1)}=\frac{1}{e^2}\lim_{x\to e}\frac{\ln x- \ln e}{e^{2\ln x-2\ln e}-1}\frac{2\ln x-2 \ln e}{2\ln x-2 \ln e}=\frac{1}{e^2}\lim_{x\to e}\frac{-(\ln x-\ln e)}{2(\ln x-\ln e)}=-\frac{1}{2e^2}
$$
$$L=e^{-\frac{1}{2e^2}}=\frac{1}{\sqrt[2e^2]{e}}$$
That's how I think this limit should be solved but my teacher got a different result.. What did I do wrong?
It seems correct indeed let $y=x-e \to 0$
$$\lim_{x\to e} (2-\ln x)^\frac{1}{x^2-e^2}=\lim_{y\to 0} (2-\ln (y+e))^\frac{1}{y(y+2e)}$$
and
$$(2-\ln (y+e))^\frac{1}{ y(y+2e) }=[(1+( 1-\ln(y+e) )^{\frac1{1-\ln(y+e)}}]^{\frac{1-\ln(y+e)}{y(y+2e)}}\to e^{-\frac1{2e^2}}$$
indeed
$$\frac{1-\ln(y+e)}{y(y+2e)}=\frac{1-\ln e-\ln(1+y/e)}{y(y+2e)}=-\frac1e \frac{\ln(1+y/e)}{y/e}\frac{1}{(y+2e)}\to -\frac1{2e^2}$$