Answer
Please see below.
Work Step by Step
As we know, we have the trigonometric relation $1-\cos 2u =2 \sin ^2 u$. By substituting $u=\frac{x}{2}$ we get $1-\cos x = 2 \sin ^2 \frac{x}{2}$.
Also, by substituting $t=\frac{x}{2}$ in Theorem 1.9(1), $\lim_{t \to 0} \frac{\sin t}{t}=1$, we get $\lim_{x \to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=1$.
So, by applying Theorem 1.2(3), $\lim_{x \to c}[f(x)g(x)]=[\lim_{x \to c}f(x)][\lim_{x \to c}g(x)]$, and Theorem 1.5, $\lim_{x \to c}f(g(x))= f \bigg(\lim_{x \to c} g(x) \bigg )$, we conclude that$$\lim_{x \to 0} \frac{1-\cos x}{x}=\lim_{x \to 0} \frac{2 \sin ^2 \frac{x}{2}}{x}=\lim_{x \to 0} \frac{\sin ^2 \frac{x}{2}}{\frac{x}{2}}= \lim_{x \to 0} \frac{x}{2} \left (\frac{ \sin \frac{x}{2}}{\frac{x}{2}} \right )^2= 0 \times 1^2=0$$(Please note that we are eligible to multiply the above fraction by $\frac{x}{x}$ since $x \neq 0$ in the definition of $\lim_{x \to 0}f(x)$).
