Is this sloppy writing for limits?

Question

Please note that I am not asking you to compute or show me how to do this limit. I am asking how to write out a clean and formal solution that is free of any error, ambiguity, or sloppiness.

Given $$\lim\limits_{(x,y) \to (0,0)} \dfrac{3x^2 y}{x^2 + y^2}$$, find its limit

So find the limit along $y = mx$ and let $f(x,y) = \dfrac{3x^2 y}{x^2 + y^2}$. So we have $f(x,mx) = \dfrac{3x^2 mx}{x^2 + m^2 x^2} = \dfrac{3mx}{1 + m^2}$

Here is the part where I am not so hot on.

Can I write this?

$\lim\limits_{(x,y) \to (0,0)} f(x,y) = \lim\limits_{(x,y) \to (0,0)} f(x,mx) = \lim\limits_{(x,y) \to (0,0)} \dfrac{3mx}{1 + m^2}= 0$

And conclude the limit is indeed $0$ through any line. (a formal justification involves epsilon-delta, but I omit it here because that is another question for another time).

I am thinking that the first equality sign is wrong.

Remark Most books I've read seem to do everything without the limit operator. Stewart for instance just argues the limit is this and this along this path and that path. I want to do my answers with the limit operators

Answer 1

There are so many ways to approach (0; 0), appoaching along the lines $y = mx$ is 1 way, other ways are approaching along the curves $y = x^2$, or $y = \sqrt{x}$, or $y = x^3$, or... any curves that pass through (0; 0).

So, pointing out that as $(x; y) \rightarrow (0; 0)$ along the lines $y = mx$ the limit is 0, is definitely not enough to show that the limit does exist.

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Here's a counter-example

Example

Evaluate $\lim\limits_{\substack{x \rightarrow 0\\y \rightarrow 0}} \dfrac{x^2 y}{x^4 + y^2}$

You can show that the limit is always 0, as $(x; y) \rightarrow (0; 0)$ along any lines $y = mx$. I'll leave this to you, it should be as easy as a piece of cake. Let's try it. :)

But if you let $(x; y) \rightarrow (0; 0)$ along the curve $y = x^2$, then you'll have: $\lim\limits_{\substack{x \rightarrow 0\\y \rightarrow 0}} \dfrac{x^2 y}{x^4 + y^2} = \lim\limits_{x \rightarrow 0} \dfrac{x^4}{x^4 + x^4} = \dfrac{1}{2}$.

So that limit does not exist.

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You can change it into polar co-ordinate, like this: Let $\left\{ \begin{array}{l} x = r \cos \varphi \\ y = r \sin \varphi \end{array} \right.$, when $(x; y) \rightarrow (0; 0)$ it means that $r \rightarrow 0$, and $\varphi$ can vary freely. So, to show that the limit exists, all you must do is to show that as $r \rightarrow 0$, and $\varphi$ takes any value, the limit stays the same.

Like this:

$\lim\limits_{\substack{x \rightarrow 0 \\ y \rightarrow 0}}\dfrac{3x^2y}{x^2 + y^2} = \lim\limits_{r \rightarrow 0}\dfrac{3r^3\cos^2 \varphi \sin \varphi}{r^2} = \lim\limits_{r \rightarrow 0} 3r\cos^2 \varphi \sin \varphi = 0$.

Of course the limit of the final expression does not depend on the value of $\varphi$, you can use the Squeeze Theorem to see this: $-3r \le 3r \cos^2 \varphi \sin \varphi \le 3r$.

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Another way to prove this is to notice that:

• $|f(x)| \rightarrow 0 \Leftrightarrow f(x) \rightarrow 0$.
• $x^2 + y^2 \ge 2 |xy|$

So, we have:

$0 \le \left| \dfrac{3x^2y}{x^2 + y^2} \right| \le \left| \dfrac{3x^2y}{2xy} \right| = \dfrac{3}{2} \left| x \right|$

Now, take the limit of the whole thing as $(x; y) \rightarrow (0; 0)$, and apply Squeeze Theorem. :)

Answer 2

Just to see how bad things can be while being totally nice along straight lines. Consider the function $f:\mathbb R^2 \to \mathbb R$ that is defined like this. For any point $(x,y)$ consider the slope of the line through the origin on which it lies (take care of the case $(x,y)=(0,0)$ separatel). Then if the slope is not of the form $1/n$, $n\in \mathbb N$ then set $f(x,y)=0$. Otherwise $(x,y)$ lies on a line through the origin of slope $1/n$ and then set $f(x,y)=0$ if the distance from $(x,y)$ to the origin is less then $1/n$, and set $f(x,y)=1$ otherwise. The limit at the origin along any straight line is $0$ though the global limit at the origin does not exist.