Using L'Hospital's Rule to evaluate limit to infinity

Question

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.

Answer 1

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).

Answer 2

Hint: Notice that $\lim_{x\to\infty}\ln\left(1+\frac{11}{x}\right)=0$, thus you might want to bring the limit in the form: $$\frac{1}{9}\lim_{x\to\infty}\frac{\ln\left(1+\frac{11}{x}\right)}{\frac{1}{x}}$$ in order to use de l'Hôpital's rule.

Answer 3

So the first step is to notice that the original equation can be simplified to $$\lim_{x \rightarrow \infty }e^{\ln{\left( \left( 1+\frac{11}{x} \right) ^{\frac{x}{9}}\right)}}$$

Which can be made into: $$e^{\lim_{x \rightarrow \infty }\frac{x}{9}(\ln{ 1+\frac{11}{x}})}$$

So now we jsut need to solve for the exponent of e:

$$\lim_{x \rightarrow \infty }\frac{x}{9}\ln({ 1+\frac{11}{x}})$$

Take out the 1/9:

$$\frac{1}{9}\lim_{x \rightarrow \infty }x\ln({ 1+\frac{11}{x}})$$

Which can now be written as:

$$\frac{1}{9}\lim_{x \rightarrow \infty }\frac{\ln({ 1+\frac{11}{x}})}{\frac{1}{x}}$$

And from here you can apply l'hopital rule

Answer 4

A different approach -

We know $\lim \limits_{x \to \infty}({1+{1\over x}})^x=e$.

So you can write your limit as $\lim \limits_{x \to \infty}({1+{11\over x}})^{x\over 9}=\lim \limits_{x \to \infty}(({1+{1\over {x\over 11}}})^{x\over 11})^{11\over9}$.

Now use limit arithmatic.

$\lim \limits_{x \to \infty}({1+{1\over {x\over 11}}})^{x\over 11}=e$, so $\lim \limits_{x \to \infty}(({1+{1\over {x\over 11}}})^{x\over 11})^{11\over9}=e^{11\over 9}$.