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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?

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

You use here this equation(wiki):

enter image description here

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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?

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

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

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$ \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}} $

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

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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.

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