[Math] Continuity of a piecewise function of two variable

continuitymultivariable-calculus

I'm given this equation:

$$
u(x,y) =
\begin{cases}
\dfrac{(x^3 – 3xy^2)}{(x^2 + y^2)}\quad& \text{if}\quad (x,y)\neq(0,0)\\
0\quad& \text{if} \quad (x,y)=(0,0).
\end{cases}
$$

It seems like L'hopitals rule has been used but I'm confused because

  1. there is no limit here it's just straight up $x$ and $y$ equals zero.
  2. if I have to invoke limit here to use Lhopitals rule, there are two variables $x$ and $y$. How do I take limit on both of them?

Best Answer

Here's one option. Write $(x,y)$ in polar form: $x = r\cos(\theta)$, $y = r\sin(\theta)$. You get:

$$u(r,\theta) = \frac{r^3\cos^{3}(\theta) - 3r^3\cos(\theta)\sin^{2}(\theta)}{r^2}$$

$$u(r,\theta) = r[\cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta)]$$

Since $\cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta)$ is continuous, it is bounded. Meaning there exists some $M > 0$ such that $-M < \cos^{3}(\theta) - 3\cos(\theta)\sin^{2}(\theta) < M$.

Then you have $-Mr < u(r,\theta) < Mr$. By the squeeze theorem:

$$\lim_{(x,y) \rightarrow (0,0)}u(x,y) = \lim_{r \rightarrow 0^{+}}u(r,\theta) = 0$$

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