Continuity of a piecewise function of two variable
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
- there is no limit here it's just straight up $x$ and $y$ equals zero.
- 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?
3 Answers
$\begingroup$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:
$$\lim_{(x,y) \rightarrow (0,0)}u(x,y) = \lim_{r \rightarrow 0^{+}}u(r,\theta) = 0$$
$\endgroup$ $\begingroup$u(0,0) can't possibly exist as it would be a division by 0. You haven't given us enough information. What is the point of this equation? What is the problem you have to solve?
$\endgroup$ 2 $\begingroup$Probably, it is a piecewise definition. $u(x,y)$ makes sense as long as both $x$ and $y$ are not $0$. It is probably just to mean that: if $(x,y)\neq (0,0)$ then $u(x,y)$ is as defined (with the formula), otherwise $u(x,y)=0$. You certainly cannot plug in $(0,0)$ to the formula. It may be that the point is chosen because there is a "hole" and that $0$ would fill in the hole, but this is not necessarily the case.
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