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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
the method! plz!
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
do you know l'hopital rule yet?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
nm silly question
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
i dont think i do yet
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
it 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
 one year ago

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
We 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.
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
i need help with finding another limit!
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
sdw:1347246353846:dw
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
this one is much easier rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you 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
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
oh ok! i get it!
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
thank you!
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
i actually need help again lolz
 one year ago

abannavong Group TitleBest ResponseYou've already chosen the best response.0
dw:1347247024415:dw
 one year ago
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