Example 3. Find $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2} y}{x^{2}+y^{2}}$.
Solution. The given function can be represented as:

$$
f(x, y)=\frac{x^{2} y}{x^{2}+y^{2}}=\frac{x y}{x^{2}+y^{2}} x .
$$

Since

$$
\left|\frac{x y}{x^{2}+y^{2}}\right| \leq \frac{1}{2}
$$

(this can be derived from the inequality $(x-y)^{2} \geq 0$ : $x^{2}-2 x y+y^{2} \geq 0$, $\left.x^{2}+y^{2} \geq 2 x y, \frac{1}{2} \geq \frac{x y}{x^{2}+y^{2}}\right)$, then

$$
|f(x, y)|=\left|\frac{x y}{x^{2}+y^{2}}\right| \cdot|x| \leq \frac{1}{2}|x|
$$

If given $\varepsilon>0$, then for $|x|<\delta$ and $\delta=2 \varepsilon$ we get

$$
|f(x, y)-0| \leq \frac{1}{2}|x|<\frac{1}{2} \cdot 2 \varepsilon=\varepsilon
$$

That is,

$$
|f(x, y)-0|<\varepsilon
$$

Therefore,

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2} y}{x^{2}+y^{2}}=0
$$