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abannavong
i need help with finding the limit if x-->0 sinx/x
answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly
do you know l'hopital rule yet?
if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book
i dont think i do yet
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
I agrea.. If you must know, google it. Or just wait a few weeks then you can do it in your head
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.
i need help with finding another limit!
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this one is much easier rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]
you will get \[\frac{2+x-2}{x(\sqrt{x+2}+\sqrt{2})}\] cancel and get \[\frac{1}{\sqrt{x+2}+\sqrt{2}}\] take the limit by replacing \(x\) by 0
i actually need help again lolz
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