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abannavong

  • 3 years ago

i need help with finding the limit if x-->0 sinx/x

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  1. anonymous
    • 3 years ago
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    answer or method? answer is 1 and it is best just to memorize it, you will use it repeatedly

  2. abannavong
    • 3 years ago
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    the method! plz!

  3. zzr0ck3r
    • 3 years ago
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    do you know l'hopital rule yet?

  4. anonymous
    • 3 years ago
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    if you want a proof that does not use derivatives, i.e. a geometric argument, it should be in any intro calc book

  5. zzr0ck3r
    • 3 years ago
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    nm silly question

  6. abannavong
    • 3 years ago
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    i dont think i do yet

  7. anonymous
    • 3 years ago
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    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

  8. zzr0ck3r
    • 3 years ago
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    I agrea.. If you must know, google it. Or just wait a few weeks then you can do it in your head

  9. zzr0ck3r
    • 3 years ago
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    agree*

  10. LolWolf
    • 3 years ago
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    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.

  11. abannavong
    • 3 years ago
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    kk!

  12. abannavong
    • 3 years ago
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    i need help with finding another limit!

  13. abannavong
    • 3 years ago
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    s|dw:1347246353846:dw|

  14. anonymous
    • 3 years ago
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    this one is much easier rationalize the numerator by multiplying top and bottom by \[\sqrt{2+x}+\sqrt{2}\]

  15. anonymous
    • 3 years ago
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    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

  16. abannavong
    • 3 years ago
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    oh ok! i get it!

  17. anonymous
    • 3 years ago
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    great

  18. abannavong
    • 3 years ago
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    thank you!

  19. anonymous
    • 3 years ago
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    yw

  20. abannavong
    • 3 years ago
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    i actually need help again lolz

  21. abannavong
    • 3 years ago
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    |dw:1347247024415:dw|

  22. LolWolf
    • 3 years ago
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    It's 0.

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