Question

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Answer 1

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$\lim_{x \to \inft3} \frac{1+cos3 \pi x}{sin^2 \pi x}=\lim_{x \to \inft3} \frac{1+ cos 2 \pi x*cos \pi x- sin2 \pi x* sin \pi x }{sin^2 \pi x} = \\ \\ =\lim_{x \to \inft3} \frac{1+ (1-2sin^2 \pi x)*cos \pi x- 2sin \pi x*cos \pi x *sin \pi x }{sin^2 \pi x} = \\ \\ =\lim_{x \to \inft3} \frac{1+ cos \pi x -2sin^2 \pi x *cos \pi x- 2sin^2 \pi x*cos \pi x }{sin^2 \pi x} = \\ \\ =\lim_{x \to \inft3} \frac{(1+ cos \pi x) -4sin^2 \pi x *cos \pi x}{sin^2 \pi x} =$

$=\lim_{x \to \inft3} \frac{1+ cos \pi x}{sin^2 \pi x} -\lim_{x \to \inft3} \frac{4sin^2 \pi x *cos \pi x}{sin^2 \pi x} = \\ \\ =\lim_{x \to \inft3} \frac{1+ cos \pi x}{1-cos^2 \pi x} -4\lim_{x \to \inft3} cos \pi x= \\ \\ =\lim_{x \to \inft3} \frac{1+ cos \pi x}{(1-cos \pi x)(1+cos \pi x )} -4cos3 \pi =\lim_{x \to \inft3} \frac{1}{1-cos \pi x} -4cos3 \pi = \\ \\ =\frac{1}{1-cos3 \pi} -4cos3 \pi = \frac{1}{1-(-1)} -4(-1)= \frac{1}{2} +4=\frac{9}{2}$