nurnabilashafiqah
lim sin^2 x/ 3x^2
x>0
using squeezing theorem .
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Goten77
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akim0001
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1/3 ?
nurnabilashafiqah
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yes ? help me to solve this question ..
Goten77
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well u can use locotoles rule
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Goten77
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Goten77
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Goten77
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nurnabilashafiqah
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the 2nd last solution .. it's blank ??
the last equation ..how it can that ?
wio
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@Goten77 "using squeezing theorem"
wio
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How about?\[ \large
\frac{-1}{ 3x^2}
\leq
\frac{\sin^2 (x)}{ 3x^2}
\leq
\frac{1}{ 3x^2}
\]
wio
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Hmm or maybe even \[
\large
0
\leq
\frac{\sin^2 (x)}{ 3x^2}
\leq
\frac{1}{ 3x^2}
\]
wio
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Hmm, no, these won't work because they're not defined at 0...
nurnabilashafiqah
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sin x . cos x become -sin^2 . cos^2 ??
nurnabilashafiqah
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wio ?? can you help me .
wio
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yes I am thinking about it.
wio
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You need to find two functions... \(g(x)\) and \(f(x)\).
The following must be true: \[
\lim_{x \to 0}g(x) = L \\
\lim_{x \to 0}f(x) = L \\
g(x) \leq \frac{\sin^2(x)}{3x^2} \leq f(x)
\]
tkhunny
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Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?
nurnabilashafiqah
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tkhunny : yes . I know about it . first rule in squeezing theorem, right ?
wio
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Oh, I got an idea now...
tkhunny
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Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?
nurnabilashafiqah
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I don't got it .
RaphaelFilgueiras
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nurnabilashafiqah
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ouh . okay . now I get it ! thank you <3
tkhunny
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How does \(\left(\dfrac{\sin(x)}{x}\right)^{2}\) compare to \(\dfrac{sin(x)}{x}\) for \(x \in (0,1)\)
nurnabilashafiqah
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just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny
wio
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\(x\in [-1, 1] \)
wio
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\(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)
RaphaelFilgueiras
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x must be in radians,this is the only restriction to aply the fundamental trygonometric limit
wio
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@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.
wio
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You need to find two functions that are equal to 1/3 at 0, one function must be greater than your function, the other must be less than your function
wio
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Use something like the fact that \[
\sin^2(x) \leq x^2
\]
wio
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That can get you an upperbound. Then you need a lower bound.
RaphaelFilgueiras
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@wio is rigth i did not see the part of squeezing
sirm3d
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\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]
RaphaelFilgueiras
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but if you are in radian you will have 0 < sinx²<x²
nurnabilashafiqah
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squeezing theorem have both less than and greater than ?
sorry, I'm new about this topic . please explain to me briefly .
RaphaelFilgueiras
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RaphaelFilgueiras
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that arc is x because in radian system we use 1 as radius
wio
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Okay so I'm thinking the best to use is: \[
\frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2}
\]
RaphaelFilgueiras
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so we have sinx<x<tgx