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I was wondering if in finding the limit of a two variables function (say, $F(x,y)$), I can choose the path by let $y=f(x)$, then find the limit in the same way of that in one variable functions.

For example, $$ \lim_{(x,y) \to (0,0)} \frac{xy}{x^2+xy+y^2} $$

(It has no limit there by choosing first $y=0$ and then $y=x$)

So I'm asking if the following procedures are correct:

Let $y=f(x)$ where $f(0)=0$ since the function passes $(0,0)$

The function then becomes: $$ \frac{xf(x)}{x^2 + xf(x) +f(x)^2} $$ Then it's an indeterminate form $[0/0]$, so I differentiate, $$ \frac{xf'(x)+f(x)}{2x + xf'(x)+f(x) +2f(x)f'(x)} $$ Then it's still $[0/0]$ so I differentiate again, $$\frac{xf''(x)+2f'(x)}{2+xf''(x)+2f'(x) +2f(x)f"(x)+2f'(x)^2}$$

By substituting $x=0$, I get $$\lim F(x,y) = \frac{2f'(x)}{2+2f'(x)+2f'(x)^2}$$

Since $f'(x)$ depends on the path I choose, the limit depends on the path I choose also. Thus, the limit at $(0,0)$ does not exist.

So that's all, the question I have are

  1. is this a valid method to determine existence of limits?
  2. is this a valid method to find the limit?

(My teacher says it wont work for 2.) but I'm still unclear about his explanations)

(Sorry if I made any mistake or this is a very stupid question, I'm very new to this site and this is my first question, thank you in advance!)

celtschk
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Sumo
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2 Answers2

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If the limit does not exist, you can find two paths that disagree. However, in case the limit does exit, your failure to find such paths is not a proof of anything. You will need to show that the limits are the same for all paths.

Karolis Juodelė
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I appreciate your efforts, this is something that every student tries to do while studying calculus of several variables. However it is a rather poor idea. Let me explain.

In the definition of limits, the two independent variables that describe your function $F$ are free to approach the limit point as they wish. Therefore any attempt to constrain them will be useful only to disprove the existence of the limit. As you know, if two different paths produce two different limits, the limit cannot exist. On the contrary, a functional relation like $y=f(x)$ will not suffice, since the point $(x,y)$ can approach another point in infinitely many other ways. Limits are characterized in terms of neighborhoods, and neighborhoods are not easily described in terms of graphs like $y=f(x)$. Although a curve can fill a square, a neighborhood is a very... thick object, while a differentiable graph is rather... thin.

To summarize: use paths only to disprove. If you need to prove, you must keep $x$ and $y$ completely independent.

Siminore
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