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Jamierox4ev3r
 one year ago
Check my work please. Not sure if I'm going in the right direction here
Jamierox4ev3r
 one year ago
Check my work please. Not sure if I'm going in the right direction here

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Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4\[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}3 }{ h } \right)\]

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4I assume that i would have to multiply both numerator and denominator with a conjugate

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4and that conjugate would be \(\large\sqrt{9+h}+3\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4alright. At least that assumption was correct

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4so from there, since i would multiply \(\sqrt{9+h}3\times \sqrt{9+h}+3\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Since the middle terms cancel out, I believe we are left with \(\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in the numerator 9+h9 in the denominator ? you are correct.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Nice! 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
 one year ago
Best ResponseYou've already chosen the best response.4\(\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}\) 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
 one year ago
Best ResponseYou've already chosen the best response.4@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
 one year ago
Best ResponseYou've already chosen the best response.0I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4at this point, it doesn't.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1assuming 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
 one year ago
Best ResponseYou've already chosen the best response.1you need to show me the original function tho

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4alright. 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
 one year ago
Best ResponseYou've already chosen the best response.4@AravindG I think I already outlined that somewhere up there :P

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0You got this :) You just need to stay confident!

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4thanks :) @Jhannybean

AravindG
 one year ago
Best ResponseYou've already chosen the best response.0Yeah you are right. The limit exists.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}3 }{ h } \right) = \frac{1}{6}}\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1there 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
 one year ago
Best ResponseYou've already chosen the best response.1I knew there was something odd LAUGHING OUT LOUD

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Alright. Good to know. \(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\) \(\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4So at that point, you can plug in 0, since it wouldn't become undefined. g'nice

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1why can't we just plug it in immediately?

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4because 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

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4At least I think that's what the prof. said oo lol

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1so what does it mean that the limit = 1/6 ?

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4why dost thou question me profusely? XD

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1fundamental concepts and analysis

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0That means at some value of x, the function will approach y at 1/6,

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4^that's the technical meaning of a limit

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4At this point, I'm not sure what you're asking

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? oo

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1just telling me what the formal definition of limit does not tell me what you understand about it, that's all

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Something to do with reimann sum right? :P lol

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4Anyhow, thank you, I'll close this for now :)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.4@nincompoop Is the reimann sum the same as the epsilondelta 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
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