need help to prove limit by using delta-epsilon

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bryanng3000

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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
 
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.
 
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 <0F68AF3B1-F4E8-4F1F-BD3E-5EFE0FDD107C.jpeg
 
In your proof(2), you say e>0 choose d=
You should say, given e>0, must find d>0 such that .....
 
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.
 
Steven G said:
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.
Click to expand...
Choosing [imath]\delta[/imath]
[math] \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[/math]
I'll leave the intermediate algebra steps, and the details of the proof to you :).
 
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BigBeachBanana said:
Choosing [imath]\delta[/imath]
[math] \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[/math]
I'll leave the intermediate algebra steps, and the details of the proof to you :).
Click to expand...
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.
 
Steven G said:
Shouldn't δ just be a function of ϵ?
Click to expand...
It is. Perhaps you didn't see the [imath]\epsilon[/imath] all the way to the right?
Steven G said:
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.
Click to expand...
Here's a 5 min Youtube video.
 
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-xo|<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.
 
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.
 
Steven G said:
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.
Click to expand...
What do you think about this? I might've made a mistake somewhere...
[math]\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}\\ [/math]
Suppose [imath]\delta<1[/imath], then [imath]|x-a|<1\implies -1< x-a < 1\implies -1+a < x <1 +a[/imath]. The fraction is maximized when
[imath]x=-1+a[/imath], it follows:
[math]\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}[/math]
So
[math]|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[/math]Choose
[math]\delta=\text{min}\left(1,\frac{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}{2}\right) [/math]
 
BigBeachBanana said:
What do you think about this? I might've made a mistake somewhere...
[math]\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}\\ [/math]
Suppose [imath]\delta<1[/imath], then [imath]|x-a|<1\implies -1< x-a < 1\implies -1+a < x <1 +a[/imath]. The fraction is maximized when
[imath]x=-1+a[/imath], it follows:
[math]\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}[/math]
So
[math]|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[/math]Choose
[math]\delta=\text{min}\left(1,\frac{|\sqrt{2a-1}|(\sqrt{2a-3}+\sqrt{2a-1})^2}{2}\right) [/math]
Click to expand...
Yes, of course. I actually knew that. The thing that through threw me off was when I used h instead of x-xo, 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.
 
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