Finding the Limit of $\frac{5^x-1}{4x}^\frac{1}{x}$

[Zhalfirin88]

Homework Statement

Find the limit of $$\lim_{x\rightarrow 0} {{\frac{(5^x-1)}{4x}^\frac{1}{x}}$$

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?

 

[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.

 

[Mark44]

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.

 

[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 =(

 

[Dick]

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 =(

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

 

[Mark44]

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.

 

[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.

 

[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:
$$\frac{ln(0)-ln(0)}{0}$$

 

[Dick]

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:
$$\frac{ln(0)-ln(0)}{0}$$

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.

 

[Zhalfirin88]

What about the exponent?

 

[Dick]

What about the exponent?

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

 

[Zhalfirin88]

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

 

[Dick]

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

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

 

[Zhalfirin88]

I don't know what you're asking.

 

[Dick]

I don't know what you're asking.

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

 

[Zhalfirin88]

edit

 

[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.

 

[vela]

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

That'll give you the wrong answer.

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

 

[Zhalfirin88]

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

 

[Dick]

That'll give you the wrong answer.

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

Why? I thought we were almost there.

 

[elect_eng]

That'll give you the wrong answer.

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

Why? Isn't it obvious that a number that is less than zero (EDIT: I meant between zero and one) raised to a power that approaches infinity will go to zero?

Doesn't taking the logarithm just confuse the situation more? Why not keep it simple.

My simple minded approach was to not even use l'Hospital, but to just look up the expansion of $$a^x$$ which to first order (with small x) is $$1+x\;{\rm ln} a$$. It all falls out pretty quickly from that. Maybe looking up in tables is frowned upon in math, but engineers are often watching the clock and looking for the quick answer.

 

[Dick]

Why? Isn't it obvious that a number that is less than zero raised to a power that approaches infinity will go to zero?

Doesn't taking the logarithm just confuse the situation more? Why not keep it simple.

My simple minded approach was to not even use l'Hospital, but to just look up the expansion of $$a^x$$ which to first order (with small x) is $$1+x\;{\rm ln} a$$. It all falls out pretty quickly from that. Maybe looking up in tables is frowned upon in math, but engineers are often watching the clock and looking for the quick answer.

I think you mean a number less than 1. Sure, that's what I thought. But don't forget x can be negative.

 

[elect_eng]

I think you mean a number less than 1. Sure, that's what I thought. But don't forget x can be negative.

Yes, quite right.

 

[vela]

Why? Isn't it obvious that a number that is less than zero raised to a power that approaches infinity will go to zero?

Yes, that's obvious, but the error occurred before this point, so it doesn't matter what that limit is.

Doesn't taking the logarithm just confuse the situation more? Why not keep it simple.

My simple minded approach was to not even use l'Hospital, but to just look up the expansion of $$a^x$$ which to first order (with small x) is $$1+x\;{\rm ln} a$$. It all falls out pretty quickly from that. Maybe looking up in tables is frowned upon in math, but engineers are often watching the clock and looking for the quick answer.

 

[vela]

Why? I thought we were almost there.

Try Mark's method. The limit comes out to be 5, not 0.

Edit: Oh, never mind. I brain-farted and was thinking of the limit as $x\rightarrow\infty$.

 

[Dick]

Try Mark's method. The limit comes out to be 5, not 0.

I really and truly think you and Mark are making a mistake here. This is f(x)=(((5^x)-1)/(4x))^(1/x), right? The limit is not 5. Nor is it 0. The limit is 0 as x->0+ and infinity as x->0-. Hence, no limit. Punch f(0.01) and f(-0.01) into your calculator if you have to. This is getting silly.

 

[elect_eng]

Try Mark's method. The limit comes out to be 5, not 0.

That doesn't seem right to me, but maybe I'm just being stupid.

Another thought occurred to me, and was triggered by Dick's comment that x can be negative. A negative exponent is an inverse, so don't we need to specify whether the limit is coming from the left or the right?

I think the answer is zero from the right and infinity from the left.

 

[Dick]

That doesn't seem right to me, but maybe I'm just being stupid.

Another thought occurred to me, and was triggered by Dick's comment that x can be negative. A negative exponent is an inverse, so don't we need to specify whether the limit is coming from the left or the right?

I think the answer is zero from the right and infinity from the left.

Wow. Somebody I can agree with!

 

[vela]

Sorry, I think you guys are right. I was thinking of $x\rightarrow\infty$.

 

[elect_eng]

Wow. Somebody I can agree with!

Yes, with some help I saw it eventually. That one was a little tricky. Hopefully, the OP will check in here before tomorrow morning's class.