Calculate: $\lim_{x \rightarrow e}\left(\frac{x}{e} \right)^\frac{1}{x-e}$ How to calcylate following limit?
$$\lim_{x \rightarrow e}\left(\frac{x}{e} \right)^\frac{1}{x-e}$$
Can we solve it without using L'Hospital?
 A: $$\lim_{x \rightarrow e}\left(\frac{x}{e} \right)^\frac{1}{x-e} = \lim_{y \rightarrow 1}\left(y \right)^\frac{1}{e(y-1)} = 
\lim_{y \rightarrow 1}(e)^\frac{ln(y)}{e(y-1)} = \\
e^{{\frac 1 e}\lim_{y \rightarrow 1}\frac{ln(y)}{y-1}} =
e^{{\frac 1 e}\lim_{z \rightarrow 0}\frac{ln(z+1)}{z}} =e^\frac 1 e
$$
You use here this equation(wiki):

A: Hint: $$\lim_{z\to0}\left(1+z\right)^{\frac1z}=e$$
Rewrite your limit in the form:
$$\lim_{x\to e}\left(\frac xe\right)^{\frac1{x-e}}=\lim_{t\to 0}\left(\frac {t+e}e\right)^{\frac1t}= \lim_{t\to 0}\left(1+\frac te\right)^{\frac ete}$$
Do you see how to continue?
A: Note that
$$ \left(\frac{x}{e}\right)^{\frac{1}{x-e}} = \exp\left( \frac{\log x - \log e}{x - e} \right). $$
Now, we identify that 
$$ \lim_{x\to e} \frac{\log x - \log e}{x - e} = \left. \frac{d \log x}{dx} \right|_{x=e} = \frac{1}{e}. $$
Therefore by continuity of the exponential function, we have
$$ \lim_{x\to e} \left(\frac{x}{e}\right)^{\frac{1}{x-e}} = e^{1/e}. $$
A: $ \lim_{x \to e} (\frac{x}{e})^{\frac{1}{x-e}} $
$ = \lim_{x \to e} (1 + \frac{x-e}{e})^{\frac{1}{x-e}} $
$ = \lim_{x \to e} (1 + \frac{1}{\frac{e}{x-e}})^{\frac{1}{x-e}} $
$ = \lim_{x \to e} ((1 + \frac{1}{\frac{e}{x-e}})^{\frac{e}{x-e}})^{\frac{1}{e}} $
$ = e^{\frac{1}{e}} $
A: According to this 
$$
\lim_{x\rightarrow e} \left(\frac xe\right)^{\frac 1{x-e}} = \exp \left(\lim_{x\rightarrow e} \frac {\ln \frac xe}{\frac 1{\frac 1{x-e}}}\right) \stackrel{L'H}{=} \exp \left(\lim_{x\rightarrow e} \frac 1x\right)  = e^{\frac 1e}
$$
A: Hint:Activate ln on your limit.
f(x)=(x/e)^(1/x-e)
lnlimf(x)=limlnf(x)=[as a result of logarithm laws)lim(ln(x)-ln(e)/(x-e). Both tend to 0 so you can actiavte l'hospital easily.
