Finding the limit of (1 - 5x)^(3/x) as x->0

1. Oct 8, 2008

duki

1. The problem statement, all variables and given/known data
lim of x->0 from above of $$(1-5x)^\frac{3}{x}$$

3. The attempt at a solution

I get to $$y=(1-5x)^\frac{3}{x}$$ but I'm stuck here.

2. Oct 8, 2008

Dick

Re: limits

What's the limit x->infinity of (1+1/x)^x? I'm guessing you know that. Can you rearrange your limit into something in that form. Try a change of variable like 1/u=-5x.

3. Oct 8, 2008

duki

Re: limits

Ok I get to this part:

$$ln y = \frac{3}{x} ln(1-5x)$$
I know the next step should be
$$\frac{\frac{3}{1-5x}\frac{d}{dx}(1-5x)}{1}$$

but why?

4. Oct 8, 2008

Dick

Re: limits

You are trying to apply l'Hopital's theorem to the log, aren't you? I didn't recognize it. Ok. The log of y is (3/x)*ln(1-5x). Write that as 3*ln(1-5x)/x. That has a 0/0 form. So you can use l'Hopital. It has the form f(x)/g(x) where f(x) and g(x) go to zero. What does l'Hopital say?

Last edited: Oct 8, 2008
5. Oct 8, 2008

duki

Re: limits

You're right, I don't =(

The limit you gave me, is it infinity?

6. Oct 8, 2008

Dick

Re: limits

No, it's e. But that's ok. If you can use l'Hopital it's easier than that. I edited my post a lot after I realized that. Reread it, ok?

7. Oct 8, 2008

duki

Re: limits

Alright, thanks.

I got $$\frac{\frac{3(-5)}{1-5x}}{x}$$ so far. Is that right?

8. Oct 8, 2008

Dick

Re: limits

According to l'Hopital, the candidate for the limit is f'(x)/g'(x), right? You took the derivative of the numerator. Why didn't you take the derivative of the denominator?

9. Oct 8, 2008

duki

Re: limits

Oh, ok. Oops.

so $$(1-5x) * (-15)$$ ?

10. Oct 8, 2008

Dick

Re: limits

Right. Now let x->0. What do you get? And remember this is ln(y). So what is y?

11. Oct 8, 2008

duki

Re: limits

ahhh OK. So does $$y = e^{-15}$$ ?

12. Oct 8, 2008

Dick

Re: limits

Yes. Now, if you want, tell me as an exercise why the limit I originally asked about, lim x->infinity (1+1/x)^x=e. It's the same thing really. Would be just for fun. If you get that you have it.

13. Oct 8, 2008

duki

Re: limits

hmm

I get to $$ln y = x ln (1+\frac{1}{x})$$ and I'm stuck. Is that 0/0? If so, how? Or is that infinity over infinity?

14. Oct 8, 2008

Dick

Re: limits

It's infinity*zero. Change it to ln(1+1/x)/(1/x). Now it's 0/0. Now you can apply l'Hopital. Find f'(x)/g'(x) and let x->infinity. It's easy. You're almost there.

15. Oct 8, 2008

duki

Re: limits

Ok. So when you differentiate you get -1 / -1. So then ln y = 1 and y = e^1 = e ?

16. Oct 8, 2008

Dick

Re: limits

Yes. You win! It's e. Remember the trick that if you see a limit like 0*infinity you can change it into infinity/infinity or 0/0 by moving a term into the numerator or denominator, whichever makes the derivatives easier. Well done.

17. Oct 8, 2008

duki

Re: limits

Awesome! Thanks.
Don't happen to have any more practice problems handy do you? :)

I have a test tomorrow on l'Hopital, and integration by simple sub, trig sub, partial fractions, and parts. I've been doing the same problems so many times, I'm worried I've just remembered the problems and not the techniques.

18. Oct 9, 2008

Dick

Re: limits

Nah. Don't have a big pool of practice problems. I'm sure you have plenty. Good luck. You did pretty well tonight once you got your wits about you. Get some sleep.

19. Oct 9, 2008

duki

Re: limits

Ok. Thanks again for your help!