A cura di: Administrator

Limite in forma indeterminata $frac{0}{0}$

$lim_{x rightarrow pi} frac{sin x}{2x-2pi} = lim_{x rightarrow pi} frac{sin x}{2(x-pi)}$

[posto $x-pi = t$] si ha

$lim_{t rightarrow 0} frac{sin(t+pi)}{2t} = -lim_{t rightarrow 0} frac{sin t}{2t} = -frac{1}{2}$