anonymous
  • anonymous
lim sin^2 x/ 3x^2 x>0 using squeezing theorem .
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Goten77
  • Goten77
|dw:1356618816515:dw|
anonymous
  • anonymous
1/3 ?
anonymous
  • anonymous
yes ? help me to solve this question ..

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Goten77
  • Goten77
well u can use locotoles rule |dw:1356631309095:dw|
Goten77
  • Goten77
|dw:1356631406328:dw|
Goten77
  • Goten77
|dw:1356631478538:dw|
Goten77
  • Goten77
|dw:1356631817966:dw||dw:1356631818210:dw|
anonymous
  • anonymous
the 2nd last solution .. it's blank ?? the last equation ..how it can that ?
anonymous
  • anonymous
@Goten77 "using squeezing theorem"
anonymous
  • anonymous
How about?\[ \large \frac{-1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
anonymous
  • anonymous
Hmm or maybe even \[ \large 0 \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
anonymous
  • anonymous
Hmm, no, these won't work because they're not defined at 0...
anonymous
  • anonymous
sin x . cos x become -sin^2 . cos^2 ??
anonymous
  • anonymous
wio ?? can you help me .
anonymous
  • anonymous
yes I am thinking about it.
anonymous
  • anonymous
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
  • tkhunny
Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?
anonymous
  • anonymous
tkhunny : yes . I know about it . first rule in squeezing theorem, right ?
anonymous
  • anonymous
Oh, I got an idea now...
tkhunny
  • tkhunny
Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?
anonymous
  • anonymous
I don't got it .
anonymous
  • anonymous
|dw:1356929342251:dw|
anonymous
  • anonymous
ouh . okay . now I get it ! thank you <3
tkhunny
  • tkhunny
How does \(\left(\dfrac{\sin(x)}{x}\right)^{2}\) compare to \(\dfrac{sin(x)}{x}\) for \(x \in (0,1)\)
anonymous
  • anonymous
just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny
anonymous
  • anonymous
\(x\in [-1, 1] \)
anonymous
  • anonymous
\(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)
anonymous
  • anonymous
x must be in radians,this is the only restriction to aply the fundamental trygonometric limit
anonymous
  • anonymous
thanks !
anonymous
  • anonymous
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
anonymous
  • anonymous
@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.
anonymous
  • anonymous
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
anonymous
  • anonymous
Use something like the fact that \[ \sin^2(x) \leq x^2 \]
anonymous
  • anonymous
That can get you an upperbound. Then you need a lower bound.
anonymous
  • anonymous
@wio is rigth i did not see the part of squeezing
sirm3d
  • sirm3d
\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]
anonymous
  • anonymous
but if you are in radian you will have 0 < sinx²
anonymous
  • anonymous
squeezing theorem have both less than and greater than ? sorry, I'm new about this topic . please explain to me briefly .
anonymous
  • anonymous
|dw:1356930269595:dw|
anonymous
  • anonymous
that arc is x because in radian system we use 1 as radius
anonymous
  • anonymous
Okay so I'm thinking the best to use is: \[ \frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2} \]
anonymous
  • anonymous
so we have sinx
anonymous
  • anonymous
huh ?

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