To me, it seems like we have to analyze an infinite number of paths. Stewart Calculus tries to invoke the idea of continuity - but how do we know continuity a priori, without examining a graph?

I know the epsilon-delta proof is one way, but some of the manipulations are highly nontrivial (at least to me).

Could somebody explain to a very ignorant kid who hasn't taken a class in real analysis why this is?


Short of doing a delta epsilon proof, you can prove that entire classes of functions are continuous. Examples of this include rational functions where they are defined, polynomials and trig functions.

Then, if you also know that the sum, product and composition of continuous functions are continuous, you are usually in quite good shape for most functions you see.

However, since you usually are given functions that aren't defined at the point of interest, often zero, common tactics include switching to polar coordinates (if at the origin) and showing that the limit is finite and independent of theta. Another way to get around the infinite path point you astutely bring up is show that for any arbitrary sequence $\vec{a_n}\rightarrow \vec{a}$, $$\lim_{n\rightarrow \infty}f(\vec{a_n})=f(\vec{a})$$

edit: Example problem for polar coordinates $$ f(x,y)=\frac{x^3}{x^2+y^2}, (x,y)\ne \vec{0}, f(0,0)=\vec{0} $$ Is f continous at the origin?

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  • $\begingroup$ Interesting. Could you elaborate more on the sequence idea, as well as the polar ideas? How would you be able to pick an arbitrary sequence? (And for that matter, what is an arbitrary sequence?) $\endgroup$ – Jerry Qu Sep 29 '16 at 1:19
  • $\begingroup$ That's the whole point! You don't pick a sequence, you prove the above identity for any sequence converging to the point in question, usually given some fact about $f$. As for polar coordinates, this is reserved for limits at 0, you just use the fact that in polar coordinates $x^2+y^2=r^2$ and the limit usually reduces to something like $\lim_{r\rightarrow 0}r\cos(\theta)$ which is easy to evaluate $\endgroup$ – qbert Sep 29 '16 at 1:24
  • $\begingroup$ Would there be any chance of showing an example of such a proof? $\endgroup$ – Jerry Qu Sep 29 '16 at 1:26
  • $\begingroup$ Sure I'll edit it in $\endgroup$ – qbert Sep 29 '16 at 1:26
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    $\begingroup$ I am getting that it is continuous, as the limit as r goes to zero of rcos^3(theta) is zero. $\endgroup$ – Jerry Qu Sep 29 '16 at 1:35

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