2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ The first and the second statements aren't equivalent... $\endgroup$
    – ASKASK
    Apr 26, 2014 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, 2014 at 22:01
  • $\begingroup$ @ASKASK Based on your simplification, what steps should I take to find the limit? $\endgroup$
    – calimat
    Apr 26, 2014 at 22:04
  • $\begingroup$ I'll post a solution $\endgroup$
    – ASKASK
    Apr 26, 2014 at 22:09

4 Answers 4

Reset to default
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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, 2014 at 21:58
  • 2
    $\begingroup$ @calimat: He uses $x/9 = (1/9)/(1/x)$. $\endgroup$
    – UnknownW
    Apr 26, 2014 at 22:04
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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, 2014 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, 2014 at 22:59
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .