anonymous
  • anonymous
(1-cosx)/x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I'm lost on what needs to be done to convert the bottom to x vs. x^2
dumbcow
  • dumbcow
is this a limit question
anonymous
  • anonymous
Oh, sorry, as x->0

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anonymous
  • anonymous
:P
dumbcow
  • dumbcow
have you learned derivatives yet?
anonymous
  • anonymous
multiply top and bottom by (1+cos(x)) and rework...
anonymous
  • anonymous
since (1+cos(x)) ->1 as x->0, it's okay (no division by 0)
anonymous
  • anonymous
is this making any sense?
anonymous
  • anonymous
Sorry, no.
jdoe0001
  • jdoe0001
yes @pgpilot326 \(\bf lim_{x\rightarrow 0} \cfrac{1-cos(x)}{x^2}\\ \cfrac{1-cos(x)}{x^2} \times \cfrac{1+cos(x)}{1+cos(x)} \implies \cfrac{1^2-cos^2(x)}{x^2+x^2cos(x)} \implies \cfrac{sin^2(x)}{x^2+x^2cos(x)}\\ \cfrac{sin^2(0)}{(0)^2+x^2cos(0)}\)
jdoe0001
  • jdoe0001
hmmm, that would still give us a 0 at the bottom :(
dumbcow
  • dumbcow
so no l'hopitals here then... also you will need the fact that \[\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1\]
anonymous
  • anonymous
first, we prpbably want one of these two forms... \[\lim_{x \rightarrow 0}\frac{ \sin x }{ x } \text{which } = 1 \text{ or } \lim_{x \rightarrow 0}\frac{ 1-\cos x }{ x } \text{which } = 0\] since factoring doesn't give us something nice \[\lim_{x \rightarrow 0}\frac{1- \cos x }{ x^2 }=\lim_{x \rightarrow 0}\left( \frac{1- \cos x }{ x } \right)\frac{ 1 }{ x }=0 \times \infty \] we want to try something else
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{1- \cos x }{ x^2 }=\lim_{x \rightarrow 0}\frac{\left(1-\cos x \right)\left( 1+\cos x \right) }{ \left( 1+\cos x \right) x^2 }=\lim_{x \rightarrow 0}\frac{\sin ^{2} x}{ \left( 1+\cos x \right) x^2 }\] \[=\lim_{x \rightarrow 0}\frac{\sin x }{ x }\frac{\sin x }{ x }\frac{1 }{ 1+ \cos x }=\lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{1 }{ 1+ \cos x }=1 \times 1 \times 1 = 1\]
anonymous
  • anonymous
have a look...
anonymous
  • anonymous
does this make sense?
anonymous
  • anonymous
kinda, would you just keep it 1-cosx, and plug in?
anonymous
  • anonymous
no, you get 0/0
anonymous
  • anonymous
?
anonymous
  • anonymous
At the end, you got 1*1*blank.
anonymous
  • anonymous
Would you make it 1*1*1/(1-cosx)
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{1 }{ 1+ \cos x }= 1 \times 1 \times 1 = 1\]
anonymous
  • anonymous
Ya, thet's what I was saying, lol, sorry.
anonymous
  • anonymous
well, 1- cos x is different from 1 + cos x as x ->0
anonymous
  • anonymous
Oh, sorry, meant +, my comp is lagging like crazy on openstudy.
anonymous
  • anonymous
it's okay... does it make sense now?
anonymous
  • anonymous
Yes, thanks!
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
I have tons of review to do after this
anonymous
  • anonymous
it's all practice and seeing many examples. then you start to get a feel for what's needed. just keep at it and ask for help if you get stuck
anonymous
  • anonymous
k, thanks.

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