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?

$$\lim_{x \rightarrow e}\left(\frac{x}{e} \right)^\frac{1}{x-e} = \lim_{y \rightarrow 1}\left(y \right)^\frac{1}{e(y-1)} = \lim_{y \rightarrow 1}(e)^\frac{ln(y)}{e(y-1)} = \\ e^{{\frac 1 e}\lim_{y \rightarrow 1}\frac{ln(y)}{y-1}} = e^{{\frac 1 e}\lim_{z \rightarrow 0}\frac{ln(z+1)}{z}} =e^\frac 1 e$$

You use here this equation(wiki):

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?

• @kalpeshmpopat: I edited my answer. – Dennis Gulko Mar 21 '13 at 9:26

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

$$\lim_{x\to e} \left(\frac{x}{e}\right)^{\frac{1}{x-e}} = e^{1/e}.$$

$\lim_{x \to e} (\frac{x}{e})^{\frac{1}{x-e}}$

$= \lim_{x \to e} (1 + \frac{x-e}{e})^{\frac{1}{x-e}}$

$= \lim_{x \to e} (1 + \frac{1}{\frac{e}{x-e}})^{\frac{1}{x-e}}$

$= \lim_{x \to e} ((1 + \frac{1}{\frac{e}{x-e}})^{\frac{e}{x-e}})^{\frac{1}{e}}$

$= e^{\frac{1}{e}}$

According to this $$\lim_{x\rightarrow e} \left(\frac xe\right)^{\frac 1{x-e}} = \exp \left(\lim_{x\rightarrow e} \frac {\ln \frac xe}{\frac 1{\frac 1{x-e}}}\right) \stackrel{L'H}{=} \exp \left(\lim_{x\rightarrow e} \frac 1x\right) = e^{\frac 1e}$$

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.