Question

Вычислить предел:
$lim_{x to 0} (1-cos)/sin^2x$

Answer 1

$lim_{x to 0} frac{1-cosx}{sin^2x} = [ frac{0}{0} ] = lim_{x to 0} frac{sinx}{2sinxcosx}= lim_{x to 0} frac{1}{2cosx}= frac{1}{2}$