Answer:
4
Step-by-step explanation:
[tex]\lim_{x \to \infty}\frac{1-cos4x}{1-cos2x}= \lim_{x \to 0} \frac{(1-cos4x)'}{(1-cos2x)'}= \lim_{x \to 0}\frac{4sin4x}{2sin2x}= \lim_{x \to 0}\frac{2*2sin2xcos2x}{sin2x}\\ = \lim_{x \to 0}4cos2x\\ =4[/tex]
{L'Hospital's rule}
I hope this helps you
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