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

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

correct

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

alright. At least that assumption was correct

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

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

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

at this point, it doesn't.

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

you need to show me the original function tho

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

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

thanks :) @Jhannybean

Yeah you are right. The limit exists.

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

thanks, jhan

I knew there was something odd LAUGHING OUT LOUD

jhan :*

Yep.

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

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

ye why is that?

why can't we just plug it in immediately?

interesting

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

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

why dost thou question me profusely? XD

fundamental concepts and analysis

I see o-o

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

^that's the technical meaning of a limit

and then what?

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

haha

Something to do with reimann sum right? :P lol

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

they are related

I see