Limit problem?

  • #1

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.
 
Last edited:

Answers and Replies

  • #2
954
117
Have you tried taking ln right from the beginning?
 
  • #3
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.
 
  • #4
LCKurtz
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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##.

[Edit] Guess I was a little slow to respond...
 
  • #5
954
117
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 [itex]ln(f(x))[/itex]. From there, you can then recover the limit of the original function [itex]f(x)[/itex]
 

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