anonymous
  • anonymous
find lim x approaches 0 of sin2x using sandwich thoeorem or limit properties
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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primeralph
  • primeralph
|dw:1377573102052:dw|
anonymous
  • anonymous
use limit property and the fact that sine is continuous you can write \[\lim_{x\to 0}\sin(2x)=\sin(\lim_{x\to 0}2x)=\sin(2_{\lim x\to 0}x)=\sin(0)=0\]
primeralph
  • primeralph
@xmoore I think that's the shrink theorem?

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anonymous
  • anonymous
that is the shrink wrap theorem i think i need a shrink
primeralph
  • primeralph
Shrink wrap? Sandwiches? Okay, that's enough for one night.
anonymous
  • anonymous
it is actually sin squared x not sin 2x
primeralph
  • primeralph
@satellite73
anonymous
  • anonymous
@primeralph @satellite73
anonymous
  • anonymous
then it is still zero, just looks different
anonymous
  • anonymous
\[\lim_{x\to 0}\sin^2(x)=\sin^2(\lim_{x\to 0}x)=0^2=0\]
anonymous
  • anonymous
or if you need another step \[\lim_{x\to 0}\sin^2(x)=\sin^2(\lim_{x\to 0}x)=\sin^2(0)=0^2=0\]
abb0t
  • abb0t
So, about that sandwich...
anonymous
  • anonymous
Thank you so much @satellite73
anonymous
  • anonymous
lol no sandwich for me @abb0t

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