1. **Problem statement:** We want to find the limit $$\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}}$$ and show that it equals zero without using polar coordinates. 2. **Recall the definition of limit in two variables:** The limit of a function $f(x,y)$ as $(x,y)$ approaches $(0,0)$ is $L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $\sqrt{x^2 + y^2} < \delta$, we have $|f(x,y) - L| < \epsilon$. 3. **Goal:** Show that for every $\epsilon > 0$, there exists $\delta > 0$ such that $$\left| \frac{xy}{\sqrt{x^2 + y^2}} - 0 \right| < \epsilon$$ whenever $\sqrt{x^2 + y^2} < \delta$. 4. **Estimate the absolute value:** $$\left| \frac{xy}{\sqrt{x^2 + y^2}} \right| = \frac{|x||y|}{\sqrt{x^2 + y^2}}$$ 5. **Use inequalities:** Since $|x| \leq \sqrt{x^2 + y^2}$ and $|y| \leq \sqrt{x^2 + y^2}$, we have $$|x||y| \leq \left(\sqrt{x^2 + y^2}\right) \left(\sqrt{x^2 + y^2}\right) = x^2 + y^2$$ 6. **Substitute back:** $$\frac{|x||y|}{\sqrt{x^2 + y^2}} \leq \frac{x^2 + y^2}{\sqrt{x^2 + y^2}} = \sqrt{x^2 + y^2}$$ 7. **Conclusion:** If we choose $\delta = \epsilon$, then whenever $\sqrt{x^2 + y^2} < \delta$, it follows that $$\left| \frac{xy}{\sqrt{x^2 + y^2}} \right| \leq \sqrt{x^2 + y^2} < \delta = \epsilon$$ which proves the limit is zero. **Final answer:** $$\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}} = 0$$