1. **State the problem:** We want to find the limit of the function $$f(x,y) = \frac{xy}{\sqrt{x^2 + y^2}}$$ as the point $$(x,y)$$ approaches $$(0,0)$$. 2. **Recall the definition of limit in two variables:** The limit $$\lim_{(x,y) \to (0,0)} f(x,y) = L$$ exists if and only if the value of $$f(x,y)$$ approaches the same number $$L$$ regardless of the path taken to approach $$(0,0)$$. 3. **Use polar coordinates:** To analyze the limit, convert to polar coordinates where $$x = r\cos\theta$$ and $$y = r\sin\theta$$, and $$r = \sqrt{x^2 + y^2}$$. As $$(x,y) \to (0,0)$$, we have $$r \to 0$$. 4. **Rewrite the function in polar form:** $$ f(r,\theta) = \frac{(r\cos\theta)(r\sin\theta)}{\sqrt{(r\cos\theta)^2 + (r\sin\theta)^2}} = \frac{r^2 \cos\theta \sin\theta}{r} = r \cos\theta \sin\theta $$ 5. **Analyze the limit:** Since $$|\cos\theta \sin\theta| \leq 1$$ for all $$\theta$$, we have $$ |f(r,\theta)| = |r \cos\theta \sin\theta| \leq r $$ As $$r \to 0$$, $$|f(r,\theta)| \to 0$$ regardless of $$\theta$$. 6. **Conclusion:** The limit is zero because the function approaches zero from all directions as $$(x,y) \to (0,0)$$. **Final answer:** $$\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}} = 0$$