Use Taylor's theorem to evaluate the limit

Question

The limit that I need to evaluate is $\lim_\limits{x-->x_0} \dfrac{\sin x-\sin x_0}{x-x_0}$.

Answer 1

HINT

If limit is at $x_0$ note that when $f(x)$ si differentiable

$$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$

If limit is at $0$ simply note that $\sin x\to 0$.

Answer 2

The Taylor serie for $\sin x$ is $$\sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...$$ So: $$\lim_{x\to 0} \frac{\sin x -x_0}{x-x_0}= \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...-(x_0-\frac{x_0^3}{3!}+\frac{x_0^5}{5!}-\frac{x_0^7}{7!}+\frac{x_0^9}{9!}-...)}{x-x_0} $$ $$=\frac{x_0-\frac{x_0^3}{3!}+\frac{x_0^5}{5!}-\frac{x_0^7}{7!}+\frac{x_0^9}{9!}-...}{x_0}$$ $$=1-\frac{x_0^2}{3!}+\frac{x_0^4}{5!}-\frac{x_0^6}{7!}+\frac{x_0^8}{9!}-...$$