What definition of "e" are you using?
In many text books, \(\displaystyle \lim_{n\to \infty}\left(1+ \frac{1}{n}\right)^n\) is used as the definition of "e".
If you have that, then setting m= 6/n changes \(\displaystyle \lim_{n\to\infty}\left(1+ \frac{6}{n}\right)^n\) to \(\displaystyle \lim_{m\to\infty}\left(1+ \frac{1}{m}\right)^{6m}\)\(\displaystyle = \lim_{m\to\infty}\left(\left(1+ \frac{1}{m}\right)^n\right)^6\).
The way you are doing it will work if you switch to a continuous variable and use "L'Hopital's rule". Since the logarithm function is continuous, you can take the logarithm first: \(\displaystyle \lim_{x\to\infty} ln(\left(1+ \frac{6}{x}\right)^x)=\)\(\displaystyle
\lim_{x\to\infty} x ln(1+ \frac{6}{x})\). Of those two factors, one goes to infinity while the other goes to 0 so write it as \(\displaystyle \lim_{x\to\infty}\frac{ln(1+\frac{6}{x})}{\frac{6}{x}}\). And now we can rewrite that as \(\displaystyle \lim_{u\to 0}\frac{ln(1+ u)}{u/6}\), taking \(\displaystyle u= 6/x\) so that we have "0/0". Using L'Hopital's rule, we take the derivative of numerator and denominator separately and look at \(\displaystyle \lim_{u\to 0}\frac{\frac{1}{1+u}}{1/6}\). That result is 1/(1/6)= 6 so that, taking L to be the original limit, we have ln(L)= 6 so that \(\displaystyle L= e^6\).
