Using L'Hospital's Rule to evaluate limit to infinity I'm given this problem and I'm not sure how to solve it. I was only ever given one example in class on using L'Hospital's rule like this, but it is very different from this particular problem. Can anyone please show me the steps to solve a problem like this?
Evaluate the limit using L'Hospital's rule if necessary
$$\lim_{ x \rightarrow \infty } \left( 1+\frac{11}{x} \right) ^{\frac{x}{9}}$$
Basically, I only know the first step:
$$\lim_{ x \rightarrow \infty } \frac{x}{9} \ln \left( 1+\frac{11}{x} \right)$$
WolframAlpha evaluates it as $e^{\frac{11}{9}}$ but I obviously have no idea how to get to that point.

Let $a=1-\frac{11}{x}$. We know that $$a^{x/9}=\exp\left ( \ln\left (a^{x/9}  \right ) \right )=\exp\left ( \frac{x}{9}\ln(a) \right )=\exp\left ( \frac{\ln(a)}{\left ( \frac{x}{9} \right )^{-1}} \right )$$
Since $$\lim_{x\to\infty}\frac{\ln(a)}{\left ( \frac{x}{9} \right )^{-1}}=\frac{1}{9}\lim_{x\to\infty}\frac{\ln(a)}{1/x}=\ldots \textrm{Use L'Hopital's rule}\ \ldots =\frac{11}{9}$$
we get the wished answer (like WolframAlpha).
