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nurnabilashafiqahBest ResponseYou've already chosen the best response.0
yes ? help me to solve this question ..
 one year ago

Goten77Best ResponseYou've already chosen the best response.1
well u can use locotoles rule dw:1356631309095:dw
 one year ago

Goten77Best ResponseYou've already chosen the best response.1
dw:1356631817966:dwdw:1356631818210:dw
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
the 2nd last solution .. it's blank ?? the last equation ..how it can that ?
 one year ago

wioBest ResponseYou've already chosen the best response.1
@Goten77 "using squeezing theorem"
 one year ago

wioBest ResponseYou've already chosen the best response.1
How about?\[ \large \frac{1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
Hmm or maybe even \[ \large 0 \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
Hmm, no, these won't work because they're not defined at 0...
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
sin x . cos x become sin^2 . cos^2 ??
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
wio ?? can you help me .
 one year ago

wioBest ResponseYou've already chosen the best response.1
yes I am thinking about it.
 one year ago

wioBest ResponseYou've already chosen the best response.1
You need to find two functions... \(g(x)\) and \(f(x)\). The following must be true: \[ \lim_{x \to 0}g(x) = L \\ \lim_{x \to 0}f(x) = L \\ g(x) \leq \frac{\sin^2(x)}{3x^2} \leq f(x) \]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Have you considered \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\)?
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
tkhunny : yes . I know about it . first rule in squeezing theorem, right ?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
I don't got it .
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1356929342251:dw
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
ouh . okay . now I get it ! thank you <3
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
How does \(\left(\dfrac{\sin(x)}{x}\right)^{2}\) compare to \(\dfrac{sin(x)}{x}\) for \(x \in (0,1)\)
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny
 one year ago

wioBest ResponseYou've already chosen the best response.1
\(\epsilon > 0,\quad x\in [0\epsilon, 0 + \epsilon] \)
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
x must be in radians,this is the only restriction to aply the fundamental trygonometric limit
 one year ago

wioBest ResponseYou've already chosen the best response.1
@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.
 one year ago

wioBest ResponseYou've already chosen the best response.1
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
 one year ago

wioBest ResponseYou've already chosen the best response.1
Use something like the fact that \[ \sin^2(x) \leq x^2 \]
 one year ago

wioBest ResponseYou've already chosen the best response.1
That can get you an upperbound. Then you need a lower bound.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
@wio is rigth i did not see the part of squeezing
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
but if you are in radian you will have 0 < sinx²<x²
 one year ago

nurnabilashafiqahBest ResponseYou've already chosen the best response.0
squeezing theorem have both less than and greater than ? sorry, I'm new about this topic . please explain to me briefly .
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1356930269595:dw
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
that arc is x because in radian system we use 1 as radius
 one year ago

wioBest ResponseYou've already chosen the best response.1
Okay so I'm thinking the best to use is: \[ \frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2} \]
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
so we have sinx<x<tgx
 one year ago
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