another limit of a sequence

**disclaimer** · February 12th 2008, 12:06 PM

Hello;

$\lim_{n\to \infty}n\left(e^{\frac{1}{n}}-1\right)$

Sorry if it's too simple, but how to calculate the limit above? All I can see right now is $\left[\infty\cdot0\right]$ .

Thanks.

**ThePerfectHacker** · February 12th 2008, 02:44 PM

Originally Posted by disclaimer

Hello;

$\lim_{n\to \infty}n\left(e^{\frac{1}{n}}-1\right)$

Sorry if it's too simple, but how to calculate the limit above? All I can see right now is $\left[\infty\cdot0\right]$ .

Thanks.

Here.

**galactus** · February 12th 2008, 02:59 PM

Here's one way.

Let $h=\frac{1}{n}, \;\ n=\frac{1}{h}$

Make the subs and change the limit to h-->0:

$\lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}$

This is a famous limit.

To show it we can express the derivative of e^x at x=0 as a limit.

$\frac{d}{dx}[e^{x}]|_{x=0}=\lim_{u\rightarrow{0}}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}=1$

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