Answer
$$\frac{2}{9}$$
Work Step by Step
Using the fact that $$\sin^2x=\frac{1}{2}(1-\cos 2x)$$ we have
\begin{align*}
\lim _{t\rightarrow 0} \frac{ 1-\cos 2t}{\sin^23t}&=\lim _{t\rightarrow 0} \frac{2\sin^2t}{\sin^23t}\\
&=\lim _{t\rightarrow 0} \frac{2}{9}\frac{\sin^2t}{t^2} \frac{ 9t^2}{\sin^23t}\\
&=\frac{2}{9}\lim _{t\rightarrow 0} \frac{\sin^2t}{t^2} \lim _{3t\rightarrow 0}\frac{ (3t)^2}{\sin^23t}\\
&=\frac{2}{9}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1$.
