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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes ? help me to solve this question ..

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Goten77 "using squeezing theorem"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How about?\[ \large \frac{1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2} \]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm, no, these won't work because they're not defined at 0...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin x . cos x become sin^2 . cos^2 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wio ?? can you help me .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes I am thinking about it.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, I got an idea now...

tkhunny
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356929342251:dw

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

tkhunny
 3 years 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 \(x \in (0,1)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\epsilon > 0,\quad x\in [0\epsilon, 0 + \epsilon] \)

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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
 3 years ago
Best ResponseYou've already chosen the best response.0Use something like the fact that \[ \sin^2(x) \leq x^2 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That can get you an upperbound. Then you need a lower bound.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@wio is rigth i did not see the part of squeezing

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but if you are in radian you will have 0 < sinx²<x²

anonymous
 3 years 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 .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356930269595:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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
 3 years ago
Best ResponseYou've already chosen the best response.0so we have sinx<x<tgx
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