Is there any intuitive way to check whether a function is continuous at a given point?

In my exams, the questions on continuity of multivariable functions are framed like "Discuss the continuity of $$f(x,y)$$ at $$(a,b)$$..." or "Is $$f$$ continuous at $$(a,b)$$...?", and likewise. If I know beforehand that the given function is discontinuous at a given point $$(a,b)$$, then I just need to find out two paths where the the value of $$\lim_{(x,y)\to (a,b)} f(x,y)$$ are different or not equal to $$f(a,b)$$. On the other hand, if I know that the function is continuous at the given point, then I can use the $$\varepsilon-\delta$$ definition of continuity to prove continuity. But since the questions don't seem to be giving much away about the continuity of the function at the given point, I'm not sure which approach should I take first while trying to solve the question.

Is there any quick and intuitive way to figure out whether a given multivariable function is continuous or not? At least in the cases where the given function is of the form $$\frac{p(x,y)}{q(x,y)}$$, where $$p$$ and $$q$$ are polynomials in $$x$$ and $$y$$? (Exceptions are fine. Just a generic and practically useful trick would do.)

For example, the function

$$f(x,y) =\begin{cases} \frac{x^{4}-y^{4}}{x^{4}+y^{4}} & (x,y)\neq (0,0) \\ 0 & (x,y)=(0,0) . \end{cases}$$ is discontinuous at $$(0,0)$$, while the function $$g(x,y) =\begin{cases} \frac{x^{2}y^{2}}{x^{2}+y^{2}} & (x,y)\neq (0,0) \\ 0 & (x,y)=(0,0) . \end{cases}$$ is continuous at $$(0,0)$$. Is there any easy way to pick this just by looking at the functions?

The same issue exists with finding a limit and proving the existence of the limit.

• If you can separate your function into sums, products, quotients, compositions etc. of continuous functions, then your function is continuous. Just be sure that it is defined at (a,b). Jan 1, 2023 at 5:49
• @DavidRaveh Know that. I've added an example to the question now. Jan 1, 2023 at 6:01
• The quick way is to look at the possible singularities, $0$ in your examples. Add the degrees of $x$ and $y$ in the numerator and denominator. The first example is $\dfrac{x^4}{x^4}=1 \to ?$ which doesn't leave room for a continuous extension, whereas the second example yields $\dfrac{x^4}{x^2}=x^2 \to 0$ which leaves a dominating $x^2$ term. So the general answer to your question is to analyze the singularities. There are usually 3 types: en.wikipedia.org/wiki/… Jan 1, 2023 at 6:14
• @Curiouserandcuriouser It looks incredibly hard to determine continuity in general. Intuitivity is probably too much to hope for, although it might be enough for many common exercises in an introductory course on calculus. As a first step, a better question should be "can you describe an algorithm that determines whether $\frac{p(x,y)}{q(x,y)}$, where $p$ and $q$ are given polynomials in $x$ and $y$ with integer coefficients is continuous at a given point with integer coordinates?" I have not proved that problem is indeed decidable, though. Jan 6, 2023 at 19:52

Since you asked for heuristics: the degree being higher in the numerator than the denominator suggests continuity; being equal or less suggests discontinuity. (This is a heuristic and not a theorem, there are trivial counter-examples like $$x/x$$.)

This is how I would proceed:

1. First of all, you only need to inspect the singularities, i.e. $$(x,y)$$ where $$q(x,y)=0$$.

2. At each singularity, $$(x_0,y_0)$$ check the function along the direction $$(u,v)\in \mathbb R^2 \setminus \{(0,0)\}$$, i.e. calculate $$\lim_{t\searrow 0} f\big((x_0,y_0) + (u,v)t\big),$$ and check if the limit is equal to $$f(x_0,y_0)$$ regardless of the choice of $$(u,v)$$. This should be very fast to do, so I don't see a reason not to do this calculation. If this condition is not satisfied, you found a contra-example. If not, it can give you an idea how to lead the proof of continuity.

3. There are functions that are continuous in all directions, but yet not continuous. For example the function $$f(x,y) = \begin{cases} 0 & \text{ if } y=0\\ 1 & \text{ if } y=x^2 \neq 0 \\ \frac{xy}{y-x^2} & \text{otherwise}. \end{cases}$$ To detect a more complicated case of discontinuity like this one could evaluate $$\lim_{t\searrow 0} f\big((x_0,y_0) + (u,v)t+(w,z) t^2 \big),$$ for any $$(u,v,w,z)\neq 0$$.

Usually the steps 1 and 2 can be done just by looking at the function and doing a brief calculation in your head. Step 3 is already more complicated, but usually you would be able to guess for which $$(u,v,w,z)$$ something "will go wrong".

I want to warn that english is not my math-primary language so some things might be expressed incorrectly. Maybe you can use the fact that norms are equivalent in $$\mathbb{R}^2$$.

1. Modify the function so you have to find if the obtained function is continuous on $$(0,0)$$.
2. Separate the numerator polynomial into a sum of more simple functions
3. Find the smallest upper bound norm $$k$$ for both numerator and demoninator (the norm $$k$$ is $$\|(x,y)\|_k = (x^k+y^k)^{1/k}$$)
4. If the numerator has "more" norms than the denominator then it is continuous.

For example in your given examples:

1. $$x^4-y^4 \leq x^4+y^4 = (\|(x,y)\|_4)^4$$ and $$f(x,y) \leq \frac{(\|(x,y)\|_4)^4}{(\|(x,y)\|_4)^4} = 1 \neq 0$$ Note that $$4$$ is the smallest norm we could find as an upper bound (because the numerator is at the power 4) so it is very likely that the function is not continuous at $$(0,0)$$
2. $$x^2y^2 \leq (\|(x,y)\|_2)^2(\|(x,y)\|_2)^2$$ and $$x^2 + y^2 = (\|(x,y)\|_2)^2$$ so $$f(x,y)\leq (\|(x,y)\|_2)^2$$ and for any norm, $$\|(x,y)\|\rightarrow 0$$ when $$(x,y)\rightarrow (0,0)$$ (note that norms are also continuous)

1. $$x^k \leq (\|(x,y)\|_k)^k$$ and $$y^k \leq (\|(x,y)\|_k)^k$$ for any $$(x,y)$$
2. For each norms it exists $$N \in \mathbb{R_+}$$ such as $$\|(x,y)\|_i\leq N\|(x,y)\|_j$$
To sum up, the idea is to find the smallest norms possible for the numerator so that it cancels out the denominator if $$f$$ is continuous and otherwise, otherwise, the numerator is the one canceled out.