Evaluate $\lim_{x\to\infty}(1 + \frac{2}{x})^x$ So I have to evaluate: $$\lim_{x\to\infty}\left(1 + \frac{2}{x}\right)^x$$

This is:

$$\lim_{x\to\infty} e^{x\ln\left(1 + \frac{2}{x}\right)}$$

And:

$$x\ln\left( 1 + \frac{2}{x} \right) = \frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}} $$

So I should apply L'Hospital rule and calculate:  $$\lim_{x\to\infty}\frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}}$$

However I'm stuck at this step and  I don't know how to move further and land at the final result ( that is, $\displaystyle{\lim_{x\to\infty}}\left(1 + \frac{2}{x}\right)^x = e^2$ )

( Alternative options are also welcome as long as they're more simple and straightforward to apply than L'Hospital rule ).
Since
$\lim_{x\to\infty}(1 + \frac{1}{x})^x
=e$,
$\lim_{x\to\infty}(1 + \frac{2}{x})^x
=\lim_{x\to\infty}((1 + \frac{2}{x})^{x/2})^2
=e^2
$.
The key trick for computing the last limit is to set $y=\frac{1}x$ then compute the limit $\lim_{y\to 0^+} \frac{\ln(1+2y)}{y}$, which is much easier using L'Hopital.

You can apply L'Hopital because this limit is of the indeterminate form $\frac{0}{0}$.