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What is the limit of the following expression ;

$$\lim_{x\rightarrow\infty}\frac{e\frac{1}{x}}{x\left[1-e^{\frac{1}{x}}\right]}$$

and

$$\lim_{x\rightarrow 0}\frac{e\frac{1}{x}}{x\left[1-e^{\frac{1}{x}}\right]}$$

I have tried with looking at many documents. Could you give the answer or give some hints about how to find the following limits ?

AlexR
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optimal control
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    Are you familiar with L'Hospital's rule? http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule – mfl Nov 05 '14 at 19:57

2 Answers2

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Notice that $1-e^{\frac1x} \sim -\frac1x$ as $x\to \infty$. Then

$$\lim_{x\to\infty}\frac{e^{\frac1x}}{x(1-e^{\frac1x})} = \lim_{x\to\infty}\frac{e^{\frac1x}}{x(-\frac1x))} = \lim_{x\to\infty}-e^{\frac1x} = -1$$

Paul
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Hint: let $z=1/x$, you get $$\lim_{x\to +\infty}\frac{e^{1/x}}{x(1-e^{1/x})}=\lim_{z\to 0} \frac{ze^{z}}{1-e^{z}}=\lim_{z\to 0}e^{z}\frac{z}{1-e^{z}}.$$ What is $\lim_{z\to 0}\frac{z}{1-e^{z}}$?

Milly
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  • I know that it is silly to say, $e^z$ goes to zero but $\frac{z}{1-e^{z}}$ goes to zero also ? because I suppose it makes $\frac{0}{0}$. – optimal control Nov 05 '14 at 20:41
  • Not really, $e^z\to 1$ as $z\to 0$. $\frac{e^z-1}{z}\to 1$ as $z\to 0$, hence $\frac{z}{1-e^z}\to -1$. – Milly Nov 05 '14 at 22:25