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Calculus1 15 Online
OpenStudy (anonymous):

Lim x->0 sin^2/x

OpenStudy (anonymous):

It's supposed to look like \[\lim_{x \rightarrow 0} \frac{ \sin^2x }{ x }\]

OpenStudy (anonymous):

Multiply by \(\dfrac{x}{x}\), then use the limit \[\lim_{x\to0}\frac{\sin ax}{ax}=1\] for \(a\not=0\).

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):

?

OpenStudy (anonymous):

I think you use L'Hopital's Rule giving you \[\lim_{x \rightarrow 0}(\frac{ 2\sin x \cos x }{ 1 })\] then evaluate limits from there. :)

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