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.

  • $\begingroup$ The first and the second statements aren't equivalent... $\endgroup$ – ASKASK Apr 26 '14 at 21:59
  • $\begingroup$ The proper simplification would be $e^{\frac{x}{9} \ln \left( 1+\frac{11}{x} \right)}$ I believe $\endgroup$ – ASKASK Apr 26 '14 at 22:01
  • $\begingroup$ @ASKASK Based on your simplification, what steps should I take to find the limit? $\endgroup$ – calimat Apr 26 '14 at 22:04
  • $\begingroup$ I'll post a solution $\endgroup$ – ASKASK Apr 26 '14 at 22:09

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).

  • $\begingroup$ I think I understand what you did, but does the last part of you solution need to be in $exp(...)$ for it to match the Wolfram answer? $\endgroup$ – calimat Apr 26 '14 at 22:45
  • $\begingroup$ Yes. In order to understand better, we get $a^{x/9}=\exp\left ( \frac{\ln(a)}{\left ( \frac{x}{9} \right )^{-1}} \right )\to \exp(11/9)$ as $x\to\infty$. $\endgroup$ – UnknownW Apr 26 '14 at 22:59

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.

  • $\begingroup$ Okay, I'm having trouble understanding what you did to get 1/9 to the outside of the limit part and 1/x on the bottom. $\endgroup$ – calimat Apr 26 '14 at 21:58
  • 2
    $\begingroup$ @calimat: He uses $x/9 = (1/9)/(1/x)$. $\endgroup$ – UnknownW Apr 26 '14 at 22:04

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


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


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