Answer
$1$
Work Step by Step
Here, in this problem, the point $P(x,y) \to O(0,0)$
This means that $D(P,O) \to 0$
Therefore, $D^2=x^2+y^2; D \to 0$
Now, $\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=\dfrac{0}{0}$, which shows the limit of Indeterminate form; thus, we will apply L-Hospital's rule:
We get $\lim\limits_{D \to 0}\dfrac{2 D\cos D^2}{2D}=\cos (0)=1$