need help to prove limit by using delta-epsilon

[bryanng3000]

given that f(x)=sqrt(2x-1)
by using limit h->0 (f(x+h)-f(x))/h
we will get f'(x)=1/sqrt(2x-1)
how to use the delta-epsilon to prove limit

 

[Steven G]

This is a help forum which means that we offer help in solving your problem. If you had read the posting guidelines you would have know how to make a post that will get you responses.
What is the delta-epsilon definition of a limit? Try using that definition and post back with your work so we can see where you are having trouble.

 

[bryanng3000]

my problem is have i have stuck at how to find out the relation between epsilon and delta because my epsilon only have l h l <0

 

[Steven G]

Try choosing δ < |h| < 1 (that 1 can be replaced with any positive number)

 

[Steven G]

In your proof(2), you say e>0 choose d=
You should say, given e>0, must find d>0 such that .....

 

[Steven G]

Can someone please finish this proof as I can't seem to do it. You can post it here or message it to me. Thanks for your time.

 

[BigBeachBanana]

Can someone please finish this proof as I can't seem to do it. You can post it here or message it to me. Thanks for your time.

Choosing $\delta$

Spoiler

$$\left|f'(a)-\frac{f(x)-f(a)}{x-a}\right|<\epsilon\\ \left|\frac{1}{\sqrt{2a-1}}-\frac{\sqrt{2x-1}-\sqrt{2a-1}}{x-a}\right|<\epsilon\\ \dots\\ \left|\frac{2(x-a)}{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1})^2}\right|<\epsilon\\ \implies \delta=\frac{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1})^2}{2}\cdot\epsilon$$

I'll leave the intermediate algebra steps, and the details of the proof to you .

 

[Steven G]

Choosing $\delta$

Spoiler

$$\left|f'(a)-\frac{f(x)-f(a)}{x-a}\right|<\epsilon\\ \left|\frac{1}{\sqrt{2a-1}}-\frac{\sqrt{2x-1}-\sqrt{2a-1}}{x-a}\right|<\epsilon\\ \dots\\ \left|\frac{2(x-a)}{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1)^2}}\right|<\epsilon\\ \implies \delta=\frac{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1})^2}{2}\cdot\epsilon$$

I'll leave the intermediate algebra steps, and the details of the proof to you .

I have two comments. Shouldn't δ just be a function of ϵ?
You changed the derivative definition that was being used to the alternate derivative definition. Can it be proven with the standard derivative definition. I assume so since you claim that you can have δ as a function of ϵ and x.
I have never seen this before and will have to think about it for a few minutes. It is unusual that I can't prove a limit.

 

[BigBeachBanana]

Shouldn't δ just be a function of ϵ?

It is. Perhaps you didn't see the $\epsilon$ all the way to the right?

You changed the derivative definition that was being used to the alternate derivative definition. Can it be proven with the standard derivative definition. I assume so since you claim that you can have δ as a function of ϵ and x.

Here's a 5 min Youtube video.

 

[Steven G]

It is. Perhaps you didn't see the $\epsilon$ all the way to the right?

Yes, I did, but as far as I see, δ is a function of ϵ and x.
I will watch the video.
Thanks for your help.

 

[Steven G]

I watch the video and understood every line. I could have done all that on my own. The problem is that the instructor never found δ in his video.
His final left hand side had f(x) in it. The left hand side should ultimately be something like |k||x-x_o|<ϵ where k is some real non-zero number. Then we simply choose δ = ϵ/|k|. That is, δ is a function of just ϵ.
In the problem we are discussing, we seem to be getting δ as a function of ϵ and x. This is where I am lost.

 

[Steven G]

The more that I think about it, δ can not be a function of x and ϵ. More importantly, δ can't depend on x.
Maybe you mistakenly had an x in your post that somehow canceled out. I too had an x in my proof that I could not get rid of and that is why I asked to see a proof.

 

[BigBeachBanana]

The more that I think about it, δ can not be a function of x and ϵ. More importantly, δ can't depend on x.
Maybe you mistakenly had an x in your post that somehow canceled out. I too had an x in my proof that I could not get rid of and that is why I asked to see a proof.

What do you think about this? I might've made a mistake somewhere...

Spoiler

$$\left|f'(a)-\frac{f(x)-f(a)}{x-a}\right|\\ \left|\frac{1}{\sqrt{2a-1}}-\frac{\sqrt{2x-1}-\sqrt{2a-1}}{x-a}\right|\\ \dots\\ \left|\frac{2(x-a)}{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1})^2}\right|=|2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2}\\$$
Suppose $\delta<1$, then $|x-a|<1\implies -1< x-a < 1\implies -1+a < x <1 +a$. The fraction is maximized when
$x=-1+a$, it follows:
$$\frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2} \le \frac{1}{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}$$
So
$$|2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2} \le |2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2} <\epsilon$$Choose
$$\delta=\text{min}\left(1,\frac{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}{2}\right)$$

 

[Steven G]

What do you think about this? I might've made a mistake somewhere...

Spoiler

$$\left|f'(a)-\frac{f(x)-f(a)}{x-a}\right|\\ \left|\frac{1}{\sqrt{2a-1}}-\frac{\sqrt{2x-1}-\sqrt{2a-1}}{x-a}\right|\\ \dots\\ \left|\frac{2(x-a)}{\sqrt{2a-1}(\sqrt{2x-1}+\sqrt{2a-1})^2}\right|=|2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2}\\$$
Suppose $\delta<1$, then $|x-a|<1\implies -1< x-a < 1\implies -1+a < x <1 +a$. The fraction is maximized when
$x=-1+a$, it follows:
$$\frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2} \le \frac{1}{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}$$
So
$$|2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2x-1}+\sqrt{2a-1})^2} \le |2|\cdot |x-a|\cdot \frac{1}{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2} <\epsilon$$Choose
$$\delta=\text{min}\left(1,\frac{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}{2}\right)$$

Yes, of course. I actually knew that. The thing that through threw me off was when I used h instead of x-x_o, there was no way that I saw to bound h in terms of x. If you note one of my previous post did suggest to let |h|<1.
Thanks for your time.