Ricordiamo la formula di prostaferesi: cos x − cos α = − 2 ⋅ sin x + α 2 ⋅ sin x − α 2 cos x - cos %alpha = - 2 cdot { sin {{x+%alpha}over 2} }cdot { sin {{x-%alpha}over 2} } applichiamola al limite: lim x → α cos x − cos α x − α = lim x → α − 2 ⋅ sin x + α 2 ⋅ sin x − α 2 x − α = − lim x → α sin x + α 2 ⋅ sin x − α 2 x − α 2 = − lim x → α sin x + α 2 ⋅ lim x → α sin x − α 2 x − α 2 lim from{x toward %alpha } {{ cos x - cos %alpha } over { x-%alpha } }= lim from{x toward %alpha } {{ - 2 cdot { sin {{x+%alpha}over 2} }cdot { sin {{x-%alpha}over 2} } } over { x-%alpha } }=-lim from{x toward %alpha } {{ { sin {{x+%alpha}over 2} }cdot { sin {{x-%alpha}over 2} } } over { {x-%alpha }over 2} }=-lim from{x toward %alpha }{ sin {{x+%alpha}over 2} }cdot lim from{x toward %alpha } {{ { sin {{x-%alpha}over 2} } } over { {x-%alpha }over 2} } nel secondo limite operiamo la sostituzione y = (x - α)/2. Dobbiamo anche cambiare il limite a cui deve tendere la y.
Infatti: lim x → α y ( x ) = lim x → α ( x − α ) 2 = 0 lim from{x toward %alpha } { y(x)}=lim from{x toward %alpha } { (x-%alpha) over 2}=0 Si ottiene: lim x → α cos x − cos α x − α = − lim x → α sin x + α 2 ⋅ lim y → 0 sin y y lim from{x toward %alpha } {{ cos x - cos %alpha } over { x-%alpha } }= -lim from{x toward %alpha }{ sin {{x+%alpha}over 2} }cdot lim from{y toward0 } {{ { sin {y} } } over { y }} Adesso ricordando il limite notevole: lim y → 0 sin y y = 1 lim from{y toward 0 } { {sin y} over y}=1 Calcoliamo il limite: lim x → α cos x − cos α x − α = − lim x → α sin x + α 2 ⋅ lim y → 0 sin y y = − sin α + α 2 = − sin α lim from{x toward %alpha } {{ cos x - cos %alpha } over { x-%alpha } }= -lim from{x toward %alpha }{ sin {{x+%alpha}over 2} }cdot lim from{y toward0 } {{ { sin {y} } } over { y }}=- sin { %alpha+%alpha}over 2=-sin %alpha