Answer
(a)$$\lim_{x \to 0}\frac{1- \cos x}{x^2}=\frac{1}{2}$$
(b)Please see below.
(c)$$\cos 0.1 \approx 0.995$$
(d)$$\cos 0.1 = 0.9950$$
Work Step by Step
(a) Since by direct substitution we get the indeterminate form $\frac{0}{0}$, we should use the trigonometric relation $\cos x= 1- 2 \sin ^2 \frac{x}{2}$ as folllows.$$\lim_{x \to 0}\frac{1- \cos x}{x^2}=\lim_{x \to 0}\frac{2\sin ^2 \frac{x}{2}}{x^2}=\lim_{x \to 0} \left ( \frac{1}{2 } \right ) \left ( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right )^2= \lim_{u \to 0} \left ( \frac{1}{2} \right ) \left ( \frac{\sin u}{u} \right )^2= ( \frac{1}{2}) (1)^2=\frac{1}{2}$$(In finding the limit we have used Thereom 1.9 (a), $\lim_{x \to 0} \frac{\sin x}{x}=1$, also $u=\frac{x}{2}$).
(b) By part (a), we can approximate $\frac{1- \cos x}{x^2}$ for $x$ near $0$ as follows.$$\frac{1- \cos x}{x^2}=\frac{1}{2} \quad \Rightarrow \quad \cos x \approx 1 - \frac{1}{2}x^2$$
(c) By part (b), we can approximate $\cos 0.1$:$$\cos 0.1 \approx 1- \frac{1}{2}(0.1)^2= 0.995$$
(d) As we can see, the results obtained from a calculator and the approximation are nearly the same.
