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

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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 $\lim\limits_{x \to 0}\dfrac{sinx}{x}=1$ $\lim\limits_{D \to 0}\dfrac{\sin D^2}{D^2}=1$.