Jamierox4ev3r
  • Jamierox4ev3r
Check my work please. Not sure if I'm going in the right direction here
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jamierox4ev3r
  • Jamierox4ev3r
\[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\]
Jamierox4ev3r
  • Jamierox4ev3r
I assume that i would have to multiply both numerator and denominator with a conjugate
anonymous
  • anonymous
correct

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Jamierox4ev3r
  • Jamierox4ev3r
and that conjugate would be \(\large\sqrt{9+h}+3\)
Jamierox4ev3r
  • Jamierox4ev3r
alright. At least that assumption was correct
Jamierox4ev3r
  • Jamierox4ev3r
so from there, since i would multiply \(\sqrt{9+h}-3\times \sqrt{9+h}+3\)
Jamierox4ev3r
  • Jamierox4ev3r
since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)
Jamierox4ev3r
  • 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
  • anonymous
in the numerator 9+h-9 in the denominator ? you are correct.
Jamierox4ev3r
  • 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
  • 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
  • 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
  • AravindG
I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.
Jamierox4ev3r
  • Jamierox4ev3r
at this point, it doesn't.
Jamierox4ev3r
  • Jamierox4ev3r
took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3
nincompoop
  • 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
  • nincompoop
you need to show me the original function tho
Jamierox4ev3r
  • 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
  • Jamierox4ev3r
@AravindG I think I already outlined that somewhere up there :P
Jhannybean
  • Jhannybean
You got this :) You just need to stay confident!
Jamierox4ev3r
  • Jamierox4ev3r
thanks :) @Jhannybean
AravindG
  • AravindG
Yeah you are right. The limit exists.
Jhannybean
  • Jhannybean
\[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}\]
nincompoop
  • 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
  • nincompoop
thanks, jhan
nincompoop
  • nincompoop
I knew there was something odd LAUGHING OUT LOUD
nincompoop
  • nincompoop
jhan :*
Jamierox4ev3r
  • 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
  • Jhannybean
Yep.
Jamierox4ev3r
  • Jamierox4ev3r
and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!
Jamierox4ev3r
  • Jamierox4ev3r
So at that point, you can plug in 0, since it wouldn't become undefined. g'nice
nincompoop
  • nincompoop
ye why is that?
nincompoop
  • nincompoop
why can't we just plug it in immediately?
Jamierox4ev3r
  • 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
  • nincompoop
interesting
Jamierox4ev3r
  • Jamierox4ev3r
At least I think that's what the prof. said o-o lol
nincompoop
  • nincompoop
so what does it mean that the limit = 1/6 ?
Jamierox4ev3r
  • Jamierox4ev3r
why dost thou question me profusely? XD
nincompoop
  • nincompoop
fundamental concepts and analysis
Jamierox4ev3r
  • Jamierox4ev3r
I see o-o
Jhannybean
  • Jhannybean
That means at some value of x, the function will approach y at 1/6,
Jamierox4ev3r
  • Jamierox4ev3r
^that's the technical meaning of a limit
nincompoop
  • nincompoop
and then what?
Jamierox4ev3r
  • Jamierox4ev3r
At this point, I'm not sure what you're asking
nincompoop
  • nincompoop
haha
Jamierox4ev3r
  • Jamierox4ev3r
Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o
nincompoop
  • nincompoop
just telling me what the formal definition of limit does not tell me what you understand about it, that's all
Jamierox4ev3r
  • Jamierox4ev3r
Something to do with reimann sum right? :P lol
Jamierox4ev3r
  • Jamierox4ev3r
Anyhow, thank you, I'll close this for now :)
Jamierox4ev3r
  • 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
  • nincompoop
they are related
Jamierox4ev3r
  • Jamierox4ev3r
I see

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