#### Answer

$1$

#### Work Step by Step

Here, we have $P(x,y) \to O(0,0)$
This implies that $D(P,O) \to 0$
Therefore, $D^2=x^2+y^2; D \to 0$
As we can see that $\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=\dfrac{0}{0}$.
Using Sandwich theorem that states if $\lim\limits_{x \to 0}\dfrac{sinx}{x}=1$
$\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=1$.