Prove that
\[
\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2}+y^{2}}{x^{2}+y^{2}}=0
\]
This mean \( s \) that for every \( \varepsilon>0 \) we can find some \( \delta>0 \) such that \( \left|\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right|<\varepsilon \) whenever \( 0<\sqrt{x^{2}+y^{2}}<\delta \).
Proof:
let \( \varepsilon>0 \) and \( \delta=\sqrt{\varepsilon} \) such that \( 0<\sqrt{x^{2}+y^{2}}<\delta \). Now,
\[
\begin{aligned}
\left|\frac{x^{2} y^{2}}{x^{2}+y^{2}}-0\right| & =\left|\frac{x^{2} y^{2}}{x^{2}+y^{2}}\right| \\
& =\frac{x^{2} y^{2}}{x^{2}+y^{2}} \\
& \leq \frac{\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} \\
& =x^{2}+y^{2} \\
& <\delta^{2} \\
& =(\sqrt{\varepsilon})^{2} \\
& =\varepsilon
\end{aligned}
\]
Therefore,
\[
\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y^{2}}{x^{2}+y^{2}}=0
\]
now, observe that
\[
\begin{array}{l}
\text { observe that } \\
f(0,0)=\frac{0^{2} 0^{2}}{0^{2}+0^{2}} \neq 0=\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y^{2}}{x^{2}+y^{2}}
\end{array}
\]
