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1/3 ?

yes ? help me to solve this question ..

well u can use locotoles rule
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the 2nd last solution .. it's blank ??
the last equation ..how it can that ?

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.

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...

I don't got it .

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ouh . okay . now I get it ! thank you <3

\(x\in [-1, 1] \)

\(\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon] \)

x must be in radians,this is the only restriction to aply the fundamental trygonometric limit

thanks !

Use something like the fact that \[
\sin^2(x) \leq x^2
\]

That can get you an upperbound. Then you need a lower bound.

\[\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}\]

but if you are in radian you will have 0 < sinx²

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that arc is x because in radian system we use 1 as radius