At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

This is as far as I got:|dw:1340139301477:dw|

Let's look at the definition:
\[\lim_{h \rightarrow a} f(x)=[f(x+h) -f(x) ]/h\]

Ok.

Your f(x) = sqrt(2x-1), and a=0. So you want to find the limit of f(x) as h approaches 0

How do I know a=0?

Ok. I think I understand; The IROC is 0 at a right. Let's continue.

Hello?

Think there is something wrong ... because you should be able to get that h canceled

So that is what you did?

Yes, that was my previous step before this.

I then cancelled out 2a and -2a.

what else cancels?

on top!

I forgot to cancel 1 and -1.

yep! :)

So that means we have
\[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\]
Correct?

yes.

Do you know what to do from here?

Yes. thank you very much.