Respuesta :

HomertheGenius
[tex] \lim_{x \to 0} \frac{sin ^{2}x }{1-cos x} = ( \frac{0}{0} )= \\ = \lim_{x \to 0} \frac{1-cos ^{2}x }{1-cosx}= \\ = \lim_{n \to 0} \frac{(1+cosx)(1-cosx)}{(1-cosx)}= \\ = \lim_{x \to 0} (1+cos x) = [/tex]
= 1 + cos 0 = 1 + 1 = 2

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