## Calculus 10th Edition

$\lim\limits_{x \to 0}\dfrac{\sin x(1-\cos x)}{x^{2}}=0$
$\lim\limits_{x \to 0}\dfrac{\sin x(1-\cos x)}{x^{2}}$ Applying direct substitution immediately will result in an indeterminate form. So, to evaluate the limit, let's rewrite the expression like this: $\lim\limits_{x \to 0}\dfrac{\sin x(1-\cos x)}{x^{2}}=\lim\limits_{x \to 0}\Big(\dfrac{\sin x}{x}\Big)\Big(\dfrac{1-\cos x}{x}\Big)=...$ Since the limit of a product is equal to the product of the limits of each of the factors, this expression becomes: $...=\Big(\lim\limits_{x \to 0}\dfrac{\sin x}{x}\Big)\Big(\lim\limits_{x \to 0}\dfrac{1-\cos x}{x}\Big)=...$ Both of these limits are special limits. We know that $\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$ and $\lim\limits_{x \to 0}\dfrac{1-\cos x}{x}=0$, so we get: $...=(1)(0)=0$