Example 3.9. Find the limit of the function $f(x, y)=\left(x^{2}+\right.$ $\left.+y^{2}\right)^{2} x^{2} y^{2}$ as $x \rightarrow 0$ and $y \rightarrow 0$.
Solution. For the calculation of the specified limit, it is more convenient to switch to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$. We obtain

$$
\begin{gathered}
\lim _{\substack{x \rightarrow 0 \\
y \rightarrow 0}}\left(x^{2}+y^{2}\right)^{2 x^{2} y^{2}}=\lim _{r \rightarrow 0}\left(r^{2}\right)^{2 r^{4} \cos ^{2} \varphi \sin ^{2} \varphi}= \\
=\lim _{r \rightarrow 0}\left(r^{2}\right)^{r^{4} \sin ^{2} 2 \varphi / 2}=e^{\lim _{r \rightarrow 0} r^{4} \sin ^{2} 2 \varphi \ln r}= \\
=e^{\sin ^{2} 2 \varphi \lim _{r \rightarrow 0} r^{4} \ln r}=e^{0}=1
\end{gathered}
$$

Note that the equality to zero of the exponent of the exponential in the considered example follows from the following considerations. The exponent of the exponential represents the product of a bounded function $\sin 2 \varphi$ by an infinitesimal quantity, since calculating the limit $\lim _{r \rightarrow 0} r^{4} \ln r$ by L'Hôpital's rule,

$$
\lim _{r \rightarrow 0} r^{4} \ln r=\lim _{r \rightarrow 0} \frac{(\ln r)^{\prime}}{\left(r^{-4}\right)^{\prime}}=\lim _{r \rightarrow 0}\left(-\frac{r^{4}}{4}\right)=0
$$

Therefore, the value of the exponent of the exponential as $r \rightarrow 0$ is zero.

A function $f(P)$ is called continuous at a point $P_{0} \in$ $D(f) \subset E^{n}$ if: 1) it is defined both at the point $P_{0}$ and in some neighborhood of it; 2) if for any $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for all points $P$ in the domain of definition of the function $D(f)$, for which $\rho\left(P, P_{0}\right)$ $>0\left(f\left(P_{0}\right)0(f(P)0$ there exists such $\delta(\varepsilon)>0$ that for any points $P_{1}, P_{2} \in X$, for which $\rho\left(P_{1}, P_{2}\right)<\delta$ the inequality $\left|f\left(P_{1}\right)-f\left(P_{2}\right)\right|<\varepsilon$ holds.

If a function $f(P)$ is continuous on a bounded closed set $X$, then it is uniformly continuous on $X$.