Answer
$1$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to 0}\dfrac{\sin^2 x}{\tan x^2}=\lim\limits_{x \to 0} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L-Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{x \to 0}\dfrac{2 \sin x \cos x}{2x \sec^2 x^2}=\lim\limits_{x \to 0}\dfrac{\sin x}{x} \cdot \lim\limits_{x \to 0}\dfrac{\cos x}{\sec^2 x^2}=1 \cdot 1=1$
