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anonymous
 4 years ago
i need help with finding the limit if x>0 sinx/x
anonymous
 4 years ago
i need help with finding the limit if x>0 sinx/x

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0do you know l'hopital rule yet?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont think i do yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is a pain to do without using calculus, so best idea is to look it up with calculus it is almost a triviality an nice graph will show it though http://www.wolframalpha.com/input/?i=sin%28x%29%2Fx

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0I agrea.. If you must know, google it. Or just wait a few weeks then you can do it in your head

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We have that: \[ \sin(x)\le x\le\tan(x)=\frac{\sin(x)}{\cos(x)} \]We find the limit as \(x\to 0\) of this, and use the squeeze theorem to show the following: Since \(\sin(x)\ne 0\), for this case: \[ \frac{\sin(x)}{\sin(x)}\le\frac{x}{\sin(x)}\le\frac{\frac{\sin(x)}{\cos(x)}}{\sin(x)}=\frac{1}{\cos(x)} \]We cancel the \(\sin(x)\) in the first term: \[ 1\le\frac{x}{\sin(x)}\le \frac{1}{\cos(x)} \]Since \(x\to 0\), we simply plug in \(x=0\) for the last expression: \[ 1\le\frac{x}{\sin(x)}\le \frac{1}{1}=1 \]By the squeeze theorem, then: \[ \lim_{x\to 0}\frac{x}{\sin(x)}=1 \] We wish to find: \[ \lim_{x\to 0}\frac{\sin(x)}{x}=\frac{1}{\lim_{x\to 0}\frac{x}{\sin(x)}}=\frac{1}{1}=1 \]Thus, we are done.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need help with finding another limit!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sdw:1347246353846:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this one is much easier rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you will get \[\frac{2+x2}{x(\sqrt{x+2}+\sqrt{2})}\] cancel and get \[\frac{1}{\sqrt{x+2}+\sqrt{2}}\] take the limit by replacing \(x\) by 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i actually need help again lolz

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1347247024415:dw
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