Check my work please. Not sure if I'm going in the right direction here

- Jamierox4ev3r

Check my work please. Not sure if I'm going in the right direction here

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- Jamierox4ev3r

\[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\]

- Jamierox4ev3r

I assume that i would have to multiply both numerator and denominator with a conjugate

- anonymous

correct

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## More answers

- Jamierox4ev3r

and that conjugate would be \(\large\sqrt{9+h}+3\)

- Jamierox4ev3r

alright. At least that assumption was correct

- Jamierox4ev3r

so from there, since i would multiply \(\sqrt{9+h}-3\times \sqrt{9+h}+3\)

- Jamierox4ev3r

since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)

- Jamierox4ev3r

Since the middle terms cancel out, I believe we are left with
\(\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}\)

- anonymous

in the numerator 9+h-9
in the denominator ?
you are correct.

- Jamierox4ev3r

Nice! So I wasn't sure if I was multiplying things correctly
oh, and I kind of forgot to close the parentheses in the denominator, but other that, it looks good to you? @surjithayer

- Jamierox4ev3r

\(\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}\)
The things in red will cancel out, so you should be left with
\(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)

- Jamierox4ev3r

@nincompoop from here, would I just plug in zero in place of h? (assuming, of course, that I hadn't taken any erroneous steps)

- AravindG

I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.

- Jamierox4ev3r

at this point, it doesn't.

- Jamierox4ev3r

took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3

- nincompoop

assuming that you've done correctly up to the last portion, all you have to do is plug in the value of h = 0 and simplify

- nincompoop

you need to show me the original function tho

- Jamierox4ev3r

alright. I've already verified that most of my steps were correct with surji,
and here is the original fxn
\(\Large\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\)

- Jamierox4ev3r

@AravindG I think I already outlined that somewhere up there :P

- Jhannybean

You got this :) You just need to stay confident!

- Jamierox4ev3r

thanks :) @Jhannybean

- AravindG

Yeah you are right. The limit exists.

- Jhannybean

\[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}\]

- nincompoop

there are few key points that you probably already learned whether or not limit may or may not exist based on the function itself without doing all of the work.
it cuts many of your work

- nincompoop

thanks, jhan

- nincompoop

I knew there was something odd LAUGHING OUT LOUD

- nincompoop

jhan :*

- Jamierox4ev3r

Alright. Good to know.
\(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)
\(\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}\)

- Jhannybean

Yep.

- Jamierox4ev3r

and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!

- Jamierox4ev3r

So at that point, you can plug in 0, since it wouldn't become undefined. g'nice

- nincompoop

ye why is that?

- nincompoop

why can't we just plug it in immediately?

- Jamierox4ev3r

because then the whole fraction would be set over 0. As long as the denominator is zero, we conclude that the limit does not exist, when in reality it does

- nincompoop

interesting

- Jamierox4ev3r

At least I think that's what the prof. said o-o lol

- nincompoop

so what does it mean that the limit = 1/6 ?

- Jamierox4ev3r

why dost thou question me profusely? XD

- nincompoop

fundamental concepts and analysis

- Jamierox4ev3r

I see o-o

- Jhannybean

That means at some value of x, the function will approach y at 1/6,

- Jamierox4ev3r

^that's the technical meaning of a limit

- nincompoop

and then what?

- Jamierox4ev3r

At this point, I'm not sure what you're asking

- nincompoop

haha

- Jamierox4ev3r

Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o

- nincompoop

just telling me what the formal definition of limit does not tell me what you understand about it, that's all

- Jamierox4ev3r

Something to do with reimann sum right? :P lol

- Jamierox4ev3r

Anyhow, thank you, I'll close this for now :)

- Jamierox4ev3r

@nincompoop
Is the reimann sum the same as the epsilon-delta limit definition?! Because if it is, then we barely went over that in class. And if they are indeed the same thing, then I do understand what you're saying

- nincompoop

they are related

- Jamierox4ev3r

I see

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