Hello;
$\displaystyle \lim_{n\to \infty}n\left(e^{\frac{1}{n}}-1\right)$
Sorry if it's too simple, but how to calculate the limit above? All I can see right now is $\displaystyle \left[\infty\cdot0\right]$.
Thanks.
Here's one way.
Let $\displaystyle h=\frac{1}{n}, \;\ n=\frac{1}{h}$
Make the subs and change the limit to h-->0:
$\displaystyle \lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}$
This is a famous limit.
To show it we can express the derivative of e^x at x=0 as a limit.
$\displaystyle \frac{d}{dx}[e^{x}]|_{x=0}=\lim_{u\rightarrow{0}}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}=1$