Question

Lim(1-cos2x)/(sin3x), x-->0

Answer 1

$\displaystyle \lim\limits_{x\to 0}\frac{1-\cos 2x}{\sin 3x}=\lim\limits_{x\to 0}\frac{2\sin^2x}{3\sin x -4\sin^3x}\\ \\ \lim\limits_{x\to 0}\frac{1-\cos 2x}{\sin 3x}=\lim\limits_{x\to 0}\frac{2\sin x}{3 -4\sin^2x}\\ \\ \lim\limits_{x\to 0}\frac{1-\cos 2x}{\sin 3x}=\frac{0}{3}\\ \\ \\ \boxed{\lim\limits_{x\to 0}\frac{1-\cos 2x}{\sin 3x}=0}$

Answer 2

A resposta segue anexa,


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Obrigado pela oportunidade 
Boa sorte, bons estudos!
SSRC - 2015 
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