Answer
$\dfrac{3}{2}$
Work Step by Step
Recall the Taylor series for $\cos x=1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....$ and $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
$\lim\limits_{x \to 0} \dfrac{\sin x^2}{1- \cos 2x}=\lim\limits_{x \to 0} \dfrac{3x^2-(\dfrac{9}{2}) x^6+(\dfrac{81}{40}) x^{10} -....}{2x^2-(2/3) x^4 +(4/45) x^6-...} $
or, $= \dfrac{ \lim\limits_{x \to 0}[3x^2-(\dfrac{9}{2}) x^6+(\dfrac{81}{40}) x^{10} -....]}{ \lim\limits_{x \to 0}[2x^2-(2/3) x^4 +(4/45) x^6-...]} $
or, $ =\dfrac{3-0+0-...}{2-0+0-.....}$
or, $\lim\limits_{x \to 0} \dfrac{\sin x^2}{1- \cos 2x}=\dfrac{3}{2}$
