Answer
$\frac{4}{3}$
Work Step by Step
We have to find $\lim\limits_{{x \to 0}} \frac{2x^2 }{3-3\cos x}$
But if we put $x = 0$ we get zero in both the numerator and the denominator.
We apply L'Hopital's Rule: $$\begin{aligned}
\lim\limits_{{x \to 0}} \frac{2x^2 }{3-3\cos x}&=\lim\limits_{{x \to 0}} \frac{4x }{3\sin x}.\end{aligned}$$ As we get zero both in numerator and denominator we apply L'Hopital's Rule again: $$\begin{aligned}
\lim\limits_{{x \to 0}} \frac{4x }{3\sin x}&=\lim\limits_{{x \to 0}} \frac{4 }{3\cos x}=\frac{4}{3}.\end{aligned}$$
