$ \displaystyle\lim_{x \to 0} \frac {sin4x}{x}\frac{(1-cos x)}{x^2} = \displaystyle\lim_{x \to 0} \frac {sin4x}{4x}\cdot 4 \cdot \frac{(1-cos x)}{x^2} = 1\cdot 4 \frac{1}{2} = 2 $
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