$ \displaystyle\lim_{x \to α} \frac{sinx - sinα}{x-α} = $
per sostituzione. $ t = x - α \; ⇒ \; x = t + α \; ⇒ \; $ Se x → α allora t → 0 $
$ \begin{aligned}\displaystyle\lim_{t \to 0} \frac{sin(t+α) - sinα}{t} &= \displaystyle\lim_{t \to 0} \frac{sintcosα + sinαcost - sinα}{t} \\ &= \displaystyle\lim_{t \to 0} \frac{sintcosα + sinα(cost - 1)}{t} \\ &= \displaystyle\lim_{t \to 0} \frac{sint}{t} cosα + \displaystyle\lim_{t \to 0} sinα\frac{(cost-1)}{t}\\ &= 1\cdot cosα + 0 \\ &= cosα \end{aligned}$