1. ## Indetermination 1^infinity

[IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
Hi there. I've found some difficulties on solving this limit:
$\displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}$

I thought of working with the function $f(x)=(\displaystyle\frac{x}{x-1})^{x+2}$
And then apply L'Hopital

This way:
$\displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}$

And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.

2. Don't go the L'Hopital route.
Look at this:
$\left( {\dfrac{x}
{{x + 1}}} \right)^{x + 2} = \left( {1 - \dfrac{1}
{{x + 1}}} \right)^x \left( {1 - \dfrac{1}
{{x + 1}}} \right)^2$

3. Originally Posted by Plato
Don't go the L'Hopital route.
Look at this:
$\left( {\dfrac{x}
{{x + 1}}} \right)^{x + 2} = \left( {1 - \dfrac{1}
{{x + 1}}} \right)^x \left( {1 - \dfrac{1}
{{x + 1}}} \right)^2$
Wouldn't it be more instructive to write it as $\displaystyle \left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)$?

4. Originally Posted by Drexel28
Wouldn't it be more instructive to write it as $\left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)$?
Not if you know that $\displaystyle
\lim _{x \to \infty } \left( {1 + \frac{a}
{{x + b}}} \right)^{cx} = e^{ac}$

5. Originally Posted by Plato
Not if you know that $\displaystyle
\lim _{x \to \infty } \left( {1 + \frac{a}
{{x + b}}} \right)^{cx} = e^{ac}$
I was thinking that if we let $y=x+1$ then $\displaystyle \left(1-\frac{1}{x+1}\right)^{x+1}=\left(1-\frac{1}{y}\right)^y\to e^{-1}$, but both are good.

6. Originally Posted by Ulysses
[IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
Hi there. I've found some difficulties on solving this limit:
$\displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}$

I thought of working with the function $f(x)=(\displaystyle\frac{x}{x-1})^{x+2}$
And then apply L'Hopital

This way:
$\displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}$

And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.
If you want to use L'Hospital:

$\displaystyle \lim_{n \to \infty}\left(\frac{n}{n-1}\right)^{n+2} = \lim_{n \to \infty}\left(1 + \frac{1}{n-1}\right)^{n+2}$

$\displaystyle = \lim_{n \to \infty}e^{\ln{\left[\left(1+\frac{1}{n-1}\right)^{n+2}\right]}}$

$\displaystyle = e^{\lim_{n\to \infty}\ln{\left[\left(1+\frac{1}{n-1}\right)^{n+2}\right]}}$

$\displaystyle = e^{\lim_{n \to \infty}(n+2)\ln{\left(1+\frac{1}{n-1}\right)}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\ln{\left(1+\frac{1}{n-1}\right)}}{(n-1)^{-1}}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{\frac{-(n-1)^{-2}}{1 + \frac{1}{n-1}}}{-(n-1)^{-2}}}$ by L'Hospital's Rule

$\displaystyle = e^{\lim_{n \to \infty}\frac{1}{1+\frac{1}{n-1}}}$

$\displaystyle = e^{\lim_{n \to \infty}\frac{n-1}{n}}$

$\displaystyle = e^{\lim_{n \to \infty}\left(1-\frac{1}{n}\right)}$

$\displaystyle =e^1$

$\displaystyle = e$.

PHEW!!!

7. Originally Posted by Ulysses
[IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
Hi there. I've found some difficulties on solving this limit:
$\displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}$

I thought of working with the function $f(x)=(\displaystyle\frac{x}{x-1})^{x+2}$
And then apply L'Hopital

This way:
$\displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) \ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}$

And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.

It appears that you have dropped the $\displaystyle \ln$ from your last expression. It should be:

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

The limit in this expression is of the form 0/0. Before applying L'Hôpital's rule, let's rewrite the numerator.
$\displaystyle e^{\displaystyle\lim_{x \to\infty}{\displaystyle \frac{\ln(\displaystyle \frac{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}$ $\displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\ln(x)-\ln(x-1)}\over{\displaystyle{{1}\over{(x+2)}}}}}\right) }$

Apply L'Hôpital's rule:

$\displaystyle\ = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\displaystyle{{1}\ over{x}}-{{1}\over{x-1}}\over{ \displaystyle-{{1}\over{(x+2)^2}}}}}\right) }$ $\displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\displaystyle{{x-1-x}\over{x(x-1)}}\over{ \displaystyle-{{1}\over{(x+2)^2}}}}}\right) }$ $\displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{(x+2)^2}\over{ x(x-1)}}\right) }=e^1=e$

It works out just fine with L'Hôpital.

8. Great work. Thank you all guys!