# Limit problem?

## Homework Statement

((x+2)/(x-1))^x, lim as x --> infinitiy

## Homework Equations

I think l'Hospital's Rule or something like that...Not sure where to begin with this one.

## The Attempt at a Solution

((x+2)/(x-1))^x is the same as saying ((x+2)^x)/((x-1)^x). Since the numerator and denominator appear to be +inf/+inf or 0/0 as x --> +inf or -inf respectively, I think it's okay to use l'Hospital's rule.

So I start with finding the derivative of the numerator first. y=(x+2)^x, so I take the natural log of both sides to get ln(y) = x*ln(x+2). I take the derivative of both sides, which gives y'/y = ln(x+2) + x/(x+2). I multiply both sides by y, which gives y'=((x+2)^x)*(ln(x+2) + x/(x+2)). Along the same logic, the derivative of the denominator is y'=((x-1)^x)*(ln(x-1) + x/(x-1)). Pairing the side computations together, you get (((x+2)^x)/((x-1)^x))*((ln(x+2) + x/(x+2))/(ln(x-1) + x/(x-1))). This doesn't really get me anywhere...

I know the answer is e^3, but I don't know why. Here's the online calculation tool I used to arrive at that answer.

EDIT

Nevermind, found a solution.

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Have you tried taking ln right from the beginning?

Have you tried taking ln right from the beginning?
Don't know what you mean by that.

It's fine though because I was able to find a solution! With WR's help, I was able to find out where I screwed up. In the end I got ln(y) = 3/((1-2t)*(1-t)), where t=1/x, and factored out 3 and then evaluated for t->0 (same as stating x->∞). This got me 3*(1/1), which is the same as stating ln(y)=3. I then raised it all from the power of e to get y=e^3, which is the right answer.

EDIT

Oh, for a weird reason I thought you wrote ln as "it". And yes, this time I did. Made things much easier.

LCKurtz
I didn't check all your steps. I would start by writing$$w =\ln y =x \ln\left (\frac{x+2}{x-1}\right ) = \frac{\ln\left (\frac{x+2}{x-1}\right)}{\frac 1 x}$$and use L'Hospital's rule on that. Once you have the limit of ##w## you will know the limit of ##y##.
Whenever you have the variable in the power term, it is usually easier to take ln of the function right from the beginning ie. we find the limit of $ln(f(x))$. From there, you can then recover the limit of the original function $f(x)$