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Consider $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ where

$$f(x,y):=\begin{cases} \frac{x^3}{x^2+y^2} & \textit{ if } (x,y)\neq (0,0) \\ 0 & \textit{ if } (x,y)= (0,0) \end{cases} $$

If one wants to show the continuity of $f$, I mainly want to show that

$$ \lim\limits_{(x,y)\rightarrow0}\frac{x^3}{x^2+y^2}=0$$

But what does $\lim\limits_{(x,y)\rightarrow0}$ mean? Is it equal to $\lim\limits_{(x,y)\rightarrow0}=\lim\limits_{||(x,y)||\rightarrow0}$ or does it mean $\lim\limits_{x\rightarrow0}\lim\limits_{y\rightarrow0}$?

If so, how does one show that the above function tends to zero?

Jonathan
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5 Answers5

4

Note that we have

$$\left|\frac{x^3}{x^2+y^2}\right|\le |x|$$

The limit as $(x,y)\to(0,0)$ is therefore $0$.

The limit $\lim_{(x,y)\to(0,0)}f(x,y)=L$ means that for all $\epsilon>0$, there exists a deleted neighborhood $N_{0,0}$ (e.g., there exists a $\delta>0$, such that $0<\sqrt{x^2+y^2}<\delta$), such that whenever $(x,y)\in N_{0,0}$, $|f(x,y)-L|<\epsilon$.

Note that the iterated limits $\lim_{x\to0}\lim_{y\to0}f(x,y$ and $\lim_{y\to0}\lim_{x\to0}f(x,y$ are not necessarily equal to each other or equal to the limit $\lim_{(x,y)\to(0,0)}f(x,y)$.

In THIS ANSWER, I referenced the Moore-Osgood Theorem, which gives sufficient conditions when the limit and the iterated limits are equal.

Mark Viola
  • 179,405
3

$\lim_\limits{(x,y)\to 0}$ likely means $\lim_\limits{(x,y)\to(0,0)}$, which means that $x$ and $y$ are both tending to $0$. One could use polar coordinates where $x=r\cos(\theta)$ and $y=r\sin(\theta)$ to obtain: $$\lim_{(x,y)\to(0,0)}\frac{x^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3\cos^3(\theta)}{r^2}=\lim_{r\to 0} r\cos^3(\theta)$$ Then note that $|\cos^3(\theta)|\leq 1~~~\forall\theta\in\Bbb R$.

Dave
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The formal definition is as follows: given a function of $n$ real variables (here $n=2$): $f(x_1,\ldots, x_n),$ we say that $$\lim_{(x_1,\ldots, x_n)\to (p_1,\ldots, p_n)}f(x_1,\ldots,x_n)=L$$ if for every $\epsilon>0$, there exists a $\delta$ sufficiently small that $$ \lvert (x_1,\ldots, x_n)-(p_1,\ldots, p_n)\rvert<\delta$$ implies that $$ \lvert f(x_1,\ldots, x_n)-L\rvert<\epsilon.$$ In your case, this reduces to showing that for every $\epsilon>0$, there exists a $\delta$ sufficiently small that $$ \lvert (x,y)\rvert<\delta$$ implies that $$ \lvert f(x,y)\rvert<\epsilon.$$ Once you've digested this definition, it is worthwhile to observe that as $(x,y)\to 0$, we have that $$ \bigg|\frac{x^3}{x^2+y^2}\bigg|\le \lvert x\rvert\to 0.$$

Alekos Robotis
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$$\lim_{(x,y)\rightarrow (0,0)}f(x,y)=L$$ means that for all $\epsilon>0$ there exists a $\delta>0$ such that $$0<\sqrt {x^2+y^2}<\delta \implies |f(x,y)-L|< \epsilon$$

In your case let $\delta=\epsilon$.

$$\left|\frac{x^3}{x^2+y^2}\right|=|x|\cdot \left|\frac{x^2}{x^2+y^2}\right| \leq |x|\cdot1=|x|\leq \sqrt{x^2+y^2}<\delta$$

Now we can see that the aforementioned implication holds.

dromastyx
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TLDR:You can intuitively think of it as $\lim_{||(x,y||\to 0}$.

This becomes more clear and formal switching to polar coordinates; if you write $$f(x,y)=f(r\cos \theta, r\sin \theta)$$ we say that $$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L$$ if for every $\theta \in [0,2\pi)$ we have $$\lim_{r\to 0} f(r\cos \theta+x_0, r\sin \theta+y_0)=L$$

The intuition behind this definition is that we want $$f(x,y)\stackrel{(x,y)\to (x_0,y_0)}{\longrightarrow} L$$ to be true if $f$ approaches $L$ getting closer to $(x_0,y_0)$, regardless of the direction from which this is happening.

Lonidard
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