# Thread: another limit of a sequence

1. ## another limit of a sequence

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.

2. 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.

3. 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.

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