prove that $sin$ is differentiable at 0, and find the derivative at 0.

Question

Question: Assume the inequality $|x-\sin(x)|\le x^2$.Prove that $\sin$ is differentiable at 0, and find the derivative at 0.

Attempt: $\lim_{x \to 0}\frac {\sin(x)-sin(0)}{x-0} = \lim_{x \to 0}\frac {\sin(x)}{x} $. I have to show that it exists, but don't know how to use the assumption in the question $|x-\sin(x)|\le x^2$. Could you give some help?

Thank you in advance.

Answer 1

$$\lim_{x \to 0}\frac {\sin(x)-\sin(0)}{x-0} = \lim_{x \to 0}\frac {\sin(x)}{x}$$

$$\bigg |\frac {\sin(x)}{x} -1\bigg| = \bigg |\frac {\sin(x)-x}{x} \bigg| \le \bigg |\frac {x^2}{x} \bigg | = |x|\to 0 $$

Thus $$ \lim_{x \to 0}\frac {\sin(x)}{x} =1 $$