Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 796: 21

$1$

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 $\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=\dfrac{0}{0}$. This shows that the limit of Indeterminate form thus, we will have to apply L-Hospital's rule Plug the limits, then we have $\lim\limits_{D \to 0}\dfrac{2 D\cos D^2}{2D}=\cos (0)=1$