TITLE: Is this sloppy writing for limits?
QUESTION [5 upvotes]: 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

REPLY [2 votes]: 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.