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

Goten77
 one year ago
Best ResponseYou've already chosen the best response.1well u can use locotoles rule dw:1356631309095:dw

Goten77
 one year ago
Best ResponseYou've already chosen the best response.1dw:1356631817966:dwdw:1356631818210:dw

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

wio
 one year ago
Best 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.

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

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

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]

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

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

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

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay 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
 one year ago
Best ResponseYou've already chosen the best response.0so we have sinx<x<tgx
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