Question

How do you find the limit of $\frac{1-\cos x}{x\sin x}$ as $x\to 0$?

Answer 1

As $x\to 0$, $\sin x\to 0$

we can rewrite the denominator as $x^2$
So need to find ${lim}_{x\to 0}\frac{1-\cos x}{x^2}$
since the result in an interdeterminate is $\frac{0}{0}$ we apply H-hospital ruls
$\frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}\left(x^2\right)}=\frac{\sin x}{2x}$
If we substitute 'approaching zero' as a formal $\frac{1}{\infty }$ we arrive at the expression
$\frac{\frac{1}{\infty }}{\frac{2}{\infty }}$
after cancelling this leaves with $\frac{1}{2}$
let $\sin x=x$ which gives
$\frac{\sin x}{2x}=\frac{x}{2x}=\frac{1}{2}$