I'm having issues with the said title. In general, I get some functions,determine the domain $D_f$ and additionally the limit points of your domain, since $f$ is said to be continuous at $x$ if $x$ isn't a limit point of the set $D_f$. The problem is always to find points in which the function can be extended so it is continuous. So I can't say oh it's not continous at one specific point therefore it can't be extended. E.g. if you had a function that can be extended to $(0,0)$ but not points of the form$(0,x)$ you have to prove the statement and that's a valid answer.

Some of the tools I've been given to prove the function is continous at a point are Heine, which proved to be not useful whenever you have a function that is a polynomial and more than one variable is in the denominator, and the usual $\epsilon -\delta$ which is pretty complicated to work out most of the time.

An additional problem is that I've been taught that at uni just a few days ago, and next week I'm already being tested in solving some functions that are even harder.

It's far easier to prove that a function can't be extended at a point, because then all you have to do is find two restrictions of the given function with different limit points, or two specific sequences that converge to the same point, but the limit point of the function value of the said sequences is not the same.

So basically, I'm more interested in figuring out a general approach to these problems, rather than specific solutions. It's like I still haven't got the slightest feeling beforehand, whether it will or wil not be extendable to certain points. And considering I'll get about 20 minutes to solve such a problem I really want to figure it out.

For example, consider the function:

$$f:R^3\rightarrow R,$$defined as $$f(x,y,z) = \frac{x^2-y^2+z^2}{x+y}$$

Now, $D_f = R^3\backslash\{(x,-x,z):x,z\in R\}$ and obviously every point in $R^3$ is a limit point of the set $D_f$.

First let's talk about $(0,0,0)$ After testing out some limits of restriction of this function, e.g. $lim_{x\rightarrow0}f(x,0,0) = 0 $ whichever such restriction I choose, e.g. $y = 0, x = 0, z= 0, x=y=z$ etc. the limit is either $0$ or doesn't exist. So my intuition tells me I either can't extend it or I can extend it to 0, for that point. But I'm not sure what to do here? $\epsilon - \delta$ for $0$ didn't yield any results and taking a sequence $(x_k,y_k,z_k)$ and plugging it into the function didn't work out well either.

Not to mention proving anything for $\{(x,-x,z):x,z\in R\}$.

Other examples include functions such as : $f:R^2 \rightarrow R, f(x,y) = \frac{e^{xy-1}}{x(x^2-y)}$

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    $\begingroup$ You wrote $D_f = R^2\backslash\{(x,-x,z):x,z\in R\}$: That should be $D_f = \mathbb R^3 \backslash\{(x,-x,z):x,z\in \mathbb R\}.$ $\endgroup$ – zhw. Nov 25 '17 at 0:55
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    $\begingroup$ As a heuristic, if you have a fraction where numerator and denominator can become zero simultaneously, you should always try to factor out terms. For your example, you'll get f(x, y, z) = x - y + z^2 / (x + y), which is well-defined on a larger domain, and continuous on its domain. $\endgroup$ – Ruben Nov 25 '17 at 21:32
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    $\begingroup$ Under general assumption, each case should be treated separately. A function of several variables can behave rather wildly, and no universal tool exists. The existence itself of a limit can be a challenging task, so that 20 minutes could be considered as an optimistic estimate of the time you have to spend to solve such problems! $\endgroup$ – Siminore Nov 27 '17 at 17:30

There's a nice sufficient condition for nonexistence of a limit in situations similar to your examples. (By "limit" I mean finite limit). To keep things simple, let $g,h$ be continuous on $\mathbb R^2,$ and let $Z_h$ be the zero set of $h.$ Then $f=g/h$ is defined and continuous on $D_f = \mathbb R^2\setminus Z_h.$ I'll assume each point of $Z_h$ is a limit point of $D_f.$

Claim 1: Suppose $(x_0,y_0)\in Z_h$ and $g(x_0,y_0)\ne 0$ Then $f$ is unbounded in $B((x_0,y_0),r) \cap D_f$ for every $r>0.$ Hence $f$ doesn't have a limit at $(x_0,y_0)$ within $D_f.$

Proof: Fix $r>0.$ Let $(x_n,y_n)\in D_f$ be a sequence converging to $(x_0,y_0).$ Then the tail end of this sequence lies in $B((x_0,y_0),r).$ Hence it is enough to show $f(x_n,y_n)$ is unbounded. Now


But $g,h$ are both continuous at $(x_0,y_0),$ so the numerator $g(x_n,y_n)\to g(x_0,y_0)\ne 0,$ while the denominator $h(x_n,y_n) \to h(x_0,y_0)=0.$ This implies $f(x_n,y_n)$ is unbounded, and we're done.

In your example $f(x,y) = \dfrac{e^{xy-1}}{x(x^2-y)},$ we see $Z_h$ equals the union of the curves $x=0, y=x^2.$ Since $e^{xy-1}$ is never $0,$ $f$ fails to have a limit at each point of $Z_h$ by Claim 1. Sometimes you can answer these questions quickly!

Here is a companion to Claim 1:

Claim 2: Suppose $(x_0,y_0)\in Z_h$ and it is the limit of a sequence $(x_n,y_n)$ in $Z_h$ such $g(x_n,y_n)\ne 0$ for all $n.$ Then the conclusion of Claim 1 holds.

Proof: By Claim 1, for each $n$ there exists $(x_n',y_n') \in D_f$ such that $|(x_n',y_n')-(x_n,y_n)| < 1/n$ and $|f(x_n',y_n')| > n.$ Because $(x_n,y_n)$ converges to $(x_0,y_0),$ the same is true of $(x_n',y_n').$ We have thus produced a sequence in $D_f$ converging to $(x_0,y_0)$ along which $f$ is unbounded. The conclusion follows.

These ideas, and the claims, extend naturally to $\mathbb R^3.$ Going to your example

$$f(x,y,z) = \dfrac{x^2-y^2+z^2}{x+y},$$

we have $Z_h = \{(x,-x,z): x,z\in \mathbb R\},$ as you found. If $z\ne 0,$ then $g(x,-x,z) \ne 0.$ Thus from Claim 1, $f$ has no limit at a point of $Z_h$ where $z\ne 0.$ What about the points $(x,-x,0)\in Z_h?$ We use Claim 2: If $(x_0,-x_0,0)$ is one of these, then $(x_0,-x_0,1/n)\to (x_0,-x_0,0),$ and each $(x_0,-x_0,1/n)$ satisfies the hypotheses of Claim 2. By Claim 2, $f$ does not have a limit at $(x_0,-x_0,0).$ We have shown $f$ fails to have a limit at each point in $Z_h.$

The strategy described above is not often mentioned in my experience, but as we've seen, it can be useful in the case where $f=g/h$ and $Z_h$ is "large" (say a curve or a surface). I'll stop here as this is the case I wanted to focus on.

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