$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.

Let y=(1+2/n)^n
ln⁡y=ln⁡〖(1+2/n)^n 〗
lim┬(n→∞)⁡ln⁡〖y= lim┬(n→∞)⁡ln⁡〖(1+2/n)^n 〗=ln⁡lim┬(n→∞)⁡y 
〖lim┬(n→∞)  n〗⁡ln⁡(1+2/n) =∞×0;hence undefined.
Proceed by using L’Hopitals Rule:
〖lim┬(n→∞)  n〗⁡ln⁡(1+2/n) =lim┬(n→∞)⁡〖ln⁡(1+2/n)/(1/n)〗
                         =lim┬(n→∞)⁡〖((-2)/(n^2 (1+2/n) ))/((-1)/n^2 )〗                             =lim┬(n→∞)⁡〖2/((1+2/n) )〗→2     
Hence ln⁡lim┬(n→∞)⁡y =2=ln⁡〖e^2 〗. Hence lim┬(n→∞)⁡〖(1+2/n)^n 〗=e^2
