Answer
$\frac{1}{2}$
Work Step by Step
We have to find $\lim\limits_{{x \to 0}} \frac{1-\cos x }{x\sin x}$.
If we put $x = 0$ we get zero in both the numerator and the denominator.
We apply L'Hopital's Rule:
$$\begin{aligned}
\lim\limits_{{x \to 0}} \frac{1-\cos x }{x\sin x}&=\lim\limits_{{x \to 0}} \frac{\sin x }{\sin x+x\cos x}.\end{aligned}$$ As we get zero both in numerator and denominator we apply L'Hopital's Rule again: $$\begin{aligned}
&\lim\limits_{{x \to 0}}\frac{\sin x }{\sin x+x\cos x}\\
&=\lim\limits_{{x \to 0}} \frac{\cos x }{2\cos x-x\sin x}\\
&=\frac{1}{2}.\end{aligned}$$
