Answer
$\dfrac{3}{2}$
Work Step by Step
The Taylor series for $\cos x $ can be defined as: $\cos x=1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....$ and the Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
We have
$\lim\limits_{x \to 0} \dfrac{\sin x^2}{1- \cos 2x}=\lim\limits_{x \to 0} \dfrac{3x^2-(9/2) x^6+(81/40) x^{10} -....}{2x^2-(2/3) x^4 +(4/45) x^6-...} \\= \dfrac{ \lim\limits_{x \to 0}[3x^2-(9/2) x^6+(81/40) x^{10} -....]}{ \lim\limits_{x \to 0}[2x^2-(2/3) x^4 +(4/45) x^6-...]} \\ =\dfrac{3-0+0-...}{2-0+0-.....}\\=\dfrac{3}{2}$
