# Limit goes to 0

1. Feb 18, 2010

### Zhalfirin88

1. The problem statement, all variables and given/known data
Find the limit of $$\lim_{x\rightarrow 0} {{\frac{(5^x-1)}{4x}^\frac{1}{x}}$$

3. The attempt at a solution
First I said y = that limit above and took the ln of both sides.

$$ln (y) = \lim_{x\rightarrow 0} {ln{{\frac{(5^x-1)}{4x}^\frac{1}{x}}}$$

But that gives you -infinity divided by 0 which isn't an indeterminate form. What do I do next?

Last edited: Feb 18, 2010
2. Feb 18, 2010

### Zhalfirin88

The power 1/x goes to the whole fraction, not just to the top. Can't figure out how to get latex to do that.

3. Feb 18, 2010

### Staff: Mentor

Like this:
$$\lim_{x\rightarrow 0} \left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}}$$

You can see my LaTeX code by double-clicking it.

Let y = the expression you're taking the limit of -- without the limit. Now take ln of both sides and work from there. Then take the limit of both sides.

4. Feb 18, 2010

### Zhalfirin88

That's what I did in step 3.

If you take the limit of $$\lim_{x\rightarrow 0} \left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}}$$

You get the $$\frac{ln(0)}{0}$$ Which isn't an indeterminate form. And that's where I'm stuck =(

5. Feb 18, 2010

### Dick

How did you get to ln(0)/0? Start by just looking at the (5^x-1)/(4x) part. Does that have a limit?

6. Feb 18, 2010

### Staff: Mentor

Dick,
I thought of that, but dismissed it, thinking it wasn't a good idea. I was thinking in terms of limits such as this:
$$\lim_{x \to 0} (1 + x) ^{1/x}$$

The technique you suggest won't work in the case of this problem, but that's because here we arrive at the indeterminate form $$[1^{\infty}]$$, but the OP's problem falls into a completely different category, so I think your suggestion is a good one.

7. Feb 18, 2010

### Dick

Hi Mark,

Right. (5^x-1)/(4x) has indeterminant form 0/0. But you can resolve that with l'Hopital. After that, it's not indeterminant at all. I'm just curious how Zhalfirin88 got to log(0)/0.

8. Feb 18, 2010

### Zhalfirin88

Okay, I took the ln of both sides. So you get:
$$ln(\left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}})$$

I rewrote that as

$$\frac{1}{x}ln\left({\frac{5^x-1}{4x}\right)$$

$$\frac{1}{x}\left(ln(5^x-1)-ln(4x)\right)$$

If you substitute 0:
$$\frac{ln(0)-ln(0)}{0}$$

9. Feb 18, 2010

### Dick

You can't 'substitute 0'. This is a limit problem. x/x is also undefined if you substitute x=0 but it has a perfectly good limit. Like I said, I think you should work on the limit of (5^x-1)/(4x) first. It has 0/0 FORM. Which means you can use l'Hopital. But the limit is NOT 0/0.

10. Feb 18, 2010

### Zhalfirin88

11. Feb 18, 2010

### Dick

Don't worry about the exponent until you figure out what the inside is doing. Trust me.

12. Feb 18, 2010

### Zhalfirin88

Sorry. $$\frac{ln(5)}{4}$$

13. Feb 18, 2010

### Dick

Right. So as x->0 your limit will look like (ln(5)/4)^(1/x). How does that behave? Be a little careful here.

14. Feb 18, 2010

### Zhalfirin88

I don't know what you're asking.

15. Feb 18, 2010

### Dick

I'm asking what's the limit of (ln(5)/4)^(1/x) as x->0. It's not indeterminant.

16. Feb 18, 2010

### Zhalfirin88

edit

Last edited: Feb 18, 2010
17. Feb 18, 2010

### Dick

Well, the log of that quantity is (1/x)*ln(ln(5)/4). You can't use l'Hopital, because that's not indeterminant. Suppose x is positive and close to zero? But you should also consider x is negative and close to zero. If it helps you to think a little more clearly, ln(5)/4~0.4.

18. Feb 18, 2010

### vela

Staff Emeritus
That'll give you the wrong answer.

You need to use the method Mark suggested to take the log and finding its limit.

19. Feb 18, 2010

### Zhalfirin88

Well I have to go to work. And it's due tomorrow morning D= Thanks for your help though.

20. Feb 18, 2010

### Dick

Why? I thought we were almost there.