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lim sin^2 x/ 3x^2 x>0 using squeezing theorem .

Mathematics
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|dw:1356618816515:dw|
1/3 ?
yes ? help me to solve this question ..

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Other answers:

well u can use locotoles rule |dw:1356631309095:dw|
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the 2nd last solution .. it's blank ?? the last equation ..how it can that ?
@Goten77 "using squeezing theorem"
How about?\[ \large \frac{-1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
Hmm or maybe even \[ \large 0 \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
Hmm, no, these won't work because they're not defined at 0...
sin x . cos x become -sin^2 . cos^2 ??
wio ?? can you help me .
yes I am thinking about it.
You need to find two functions... \(g(x)\) and \(f(x)\). The following must be true: \[ \lim_{x \to 0}g(x) = L \\ \lim_{x \to 0}f(x) = L \\ g(x) \leq \frac{\sin^2(x)}{3x^2} \leq f(x) \]
Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?
tkhunny : yes . I know about it . first rule in squeezing theorem, right ?
Oh, I got an idea now...
Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?
I don't got it .
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ouh . okay . now I get it ! thank you <3
How does \(\left(\dfrac{\sin(x)}{x}\right)^{2}\) compare to \(\dfrac{sin(x)}{x}\) for \(x \in (0,1)\)
just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny
\(x\in [-1, 1] \)
\(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)
x must be in radians,this is the only restriction to aply the fundamental trygonometric limit
thanks !
http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_03_lim_of_trig_func.htm
@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.
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
Use something like the fact that \[ \sin^2(x) \leq x^2 \]
That can get you an upperbound. Then you need a lower bound.
@wio is rigth i did not see the part of squeezing
\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]
but if you are in radian you will have 0 < sinx²
squeezing theorem have both less than and greater than ? sorry, I'm new about this topic . please explain to me briefly .
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that arc is x because in radian system we use 1 as radius
Okay so I'm thinking the best to use is: \[ \frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2} \]
so we have sinx
huh ?

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