Chapter 9.10, Problem 30ES

The limit of an indeterminate form as $x\to x_0$ can sometimes be found by expanding the functions involved in Taylor series about $x=x_0$ and taking the limit of the series term by term. Use this method to find the limits in these exercises.

$\begin{array}{l}\left(\text{a}\right)\underset{x\to 0}{\lim }\frac{1-\cos x}{\sin x}\\ \left(\text{b}\right)\underset{x\to 0}{\lim }\frac{\ln \sqrt{1+x}-\sin 2x}{x}\end{array}$