How to evaluate $\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{3x}$ using L'Hôpital's rule? I'm stuck on how to evaluate the following using L'Hôpital's rule:
$$\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{3x}$$
This is a problem that I encountered on Khan Academy and I attempted to understand it using the resources there.  Here are the tips given for the problem; the portion that I'm having trouble understanding is highlighted:

I also attempted to use this video  (screenshot following) to help; I understand the concepts in the video but it seems like there are some missing steps in the tips above.

I also attempted to use WolframAlpha's step-by-step solution but it was indecipherable to me.
Any help is greatly appreciated.
 A: $(1+{2 \over n})^{3n} = ((1+{2 \over n})^{n})^3$.
We have $\lim_{n \to \infty} (1+ {\alpha \over n})^n = e^\alpha$.
To see the latter using l'Hôpital, let $a_n = (1+ {\alpha \over n})^n$.
Then $\log a_n = n \log(1+ {\alpha \over n})= { \log(1+ {\alpha \over n}) \over {1 \over n}}$.
Note that $\lim_{n \to \infty} { \log(1+ {\alpha \over n}) \over {1 \over n}} = \lim_{x \to 0} { \log(1+ \alpha x) \over x}$. Using l'Hôpital, we see that the limit is $\alpha$, so we have $\lim_{n \to \infty} \log a_n = \alpha$,from which we get $\lim_{n \to \infty} a_n = e^\alpha$.
A: Let $y = \lim_{x→\infty}(1+\frac{2}{x} )^{3x} $
$\log y = \lim_{x→\infty}(3x) ~~\log~(1+\frac{2}{x} ) $
$= \lim_{x→\infty} \frac{ \log~(1+\frac{2}{x} )}{\frac{1}{3x} }$
$= 0/0 $ form.
Differentiate numerator and denominator and get limit value as
$\log y = \lim_{x→\infty} {\frac {1} {(1+2/x)} }. (\frac {-2}{x^2}).(-\frac {x^2}{3})$
=$2/3$
Hence, $y=e^{2/3}$
A: Generally when l'hospital is involved in a complex limit, you need to set the limit equal to y then apply ln to both sides. To obtain the indeterminite form you need to move a variable to the denominator. The problem usually proceeds smoothly from there.  When you do solve the limit remember that it is set to ln (y), so solving for y is the final step involved.
