I must calculate $$\lim_{t→1} \frac{\sin (t) − \sin (1)}{t − 1}$$
I can calculate it with L'Hospital's rule but I do not understand how to do it with taylor polynom. I know sin is : $$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}\quad = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\$$ But how do I use it here? Thank you.
Well, by Taylor's formula you know that
$$ \sin t = \sin 1 + \cos(1)(t-1) - \frac{\sin \xi_t}{2!} (t-1)^2, \quad \xi_t \in (1,t). $$
Back to the limit,
\begin{align*} \lim_{t\to 1} \frac{\sin t - \sin 1}{t-1} = &\lim_{t\to 1}\frac{\sin 1 + \cos (1) (t-1) - \frac{\sin \xi_t}{2!} (t-1)^2 - \sin 1}{t-1}\\ =& \cos 1 + \lim_{t \to 1}\frac{\sin \xi_t (t-1)}{2!} =\cos 1 \end{align*}