continuity of a multivariable function2

Question

I'm studying the continuity of the function $$ f(x,y) = \left\{ \begin{array}{l l} \frac{x^2y^2}{x^2+y^2} & \quad , \quad(x,y)\neq(0,0)\\ 0 & \quad , \quad(x,y)=(0,0) \end{array} \right.$$

in the point $(x,y)=(0,0)$.

It's clear to me that if a function is not continuous I have to find a case of discontinuity, but perhaps it's more difficult to prove the continuity in which I can't find a fault in the behaviour of the function.

In this case how can I show the limit $ \lim_{(x,y)\rightarrow (0,0)} f(x,y) = \lim_{(x,y)\rightarrow (0,0)} \frac{x^2y^2}{x^2+y^2} =0$?

It's simple to prove that on a line $y=mx$ through the origin $(0,0)$ but in the most general way?

Answer 1

For $x=0$ or $y=0$, it is clear that the fraction equals $0$, so we may, without loss of generality, assume that we approach the origin through a path where $x,y\neq0$. Then we can apply AM-GM to observe that $x^2+y^2\geq2\sqrt{x^2y^2}=2|xy|$. This yields $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^2y^2}{x^2+y^2}\right|\leq\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^2y^2}{2|xy|}\right|=\lim_{(x,y)\rightarrow (0,0)}\frac{|xy|}{2}=0.$$

This is just a nice trick that often works. It can be done way simpler in this case, however: $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^2y^2}{x^2+y^2}\right|\leq\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^2y^2}{x^2}\right|=\lim_{(x,y)\rightarrow (0,0)} |y^2|=0.$$

There are many strategies to tackle these problems, and it is probably best to try the second approach first. AM-GM is a nice trick that often works.

It is really important, however, that the fact that the iterated limit, for an $m\in\mathbb{R}$ $$\lim_{x\to0}\lim_{y\to mx}f(x,y)=L,$$ does not imply the limit exists. For example, the limit of $f(x,y)=\frac{x^2y}{x^4+y^2}$ as $(x,y)\to(0,0)$ exists along every straight line approaching the origin but not along the parabola $y=x^2$.

Answer 2

Consider $\varepsilon>0$. for $x,y)\neq (0,0)$

$$ f(x,y) = \frac{x^2y^2}{\|\vec x\|^2} \leq \frac{\|\vec x\|^4}{\|\vec x\|^2} = \|\vec x\|^2. $$

Thus, for any $0<\delta^2 <\varepsilon$

$$ \sqrt{x^2+y^2}<\delta \Longrightarrow |f(x,y)|\leq \delta^2<\varepsilon . $$

Answer 3

An option:

$x=r\cos t$ ; $y =r\sin t$, $r$ , real, positive; and $0 \le t <2π.$

$F(r,t):= \dfrac{r^4\sin ^2 t \cos^2 t}{r^2}=$

$r^2 \sin^2 t \cos^2 t=(1/4)r^2 \sin ^2 (2t).$

Now consider the limit $ r \rightarrow 0$.

Answer 4

You might consider a slightly more general case:

Claim. Let $\displaystyle f(x) = \frac{p(x)}{|x|^{r}}$ be an application of $\mathbb{R}^{n}$ to $\mathbb{R}$ (i.e, $x = (x_{1},...,x_{n}$)), with $p(x)$ a monomial in $m$ variables ($n\geq m$) such $\deg (p(x)) > r$. Then $$\lim_{x\to 0}f(x) = 0.$$

Proof. For each $i = 1,...,m$ we have $x_{i}^{2} \leq |x|^2$ and so $(x_{i}^{k})^{2} \leq |x|^{2k}$. Thus $$(x_{1}^{s_{1}}\cdot \ldots \cdot x_{m}^{s_{m}})^{2} \leq (|x|^{s_{m}}\cdot \ldots \cdot |x|^{s_{m}})^{2} = |x|^{2ms_{m}} \Longrightarrow \frac{(x_{1}^{s_{m}}\cdot \ldots \cdot x_{m}^{s_{m}})^{2}}{|x|^{2r}}\leq |x|^{2(ms_{m}-r)}$$ and, since $\deg (p(x)) > r$, so $ms_{m}-r\geq0$ (suppose $s_{m} = \max s_{i}$), then $$\frac{(x_{1}^{s_{m}}\cdot \ldots \cdot x_{m}^{s_{m}})^{2}}{|x|^{2r}}\leq |x|^{2(ms_{m}-r)} \Longrightarrow |f(x)| \leq |x|^{ms_{m}-r}.$$ Therefore $$\lim_{x \to 0}|f(x)| \leq \lim_{x \to 0}|x|^{ms_{m}-r} = 0.$$

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In your case, $p(x)=x^2 y^2$ has degree $4$ and if $z = (x,y)$, so $x^2 + y^2 = (\sqrt{x^2 + y^2})^{2} = |z|^{2}$ and $\deg(p(x)) = 4 > r = 2$.

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I think this works.