now, if 1/x2 IS continuous at 0,
\lim_{x \to 0} f(x) = f(0) which is a problem, because f(0) doesn't exist.
but let's go even further, let's prove that there isn't ANY real number we can pick for f(0) that will make 1/x2 continuous at 0.
now, for this limit L to exist, we need to be able to find a δ > 0, so that:
0 < |x| < δ implies |1/x2 - L| < ε.
can L be 0? choose ε = 1. that would mean that we could find δ > 0 so that 0 < |x| < δ implies |1/x2| < 1. here, we have another problem.
|1/x2| = (1/|x|)2, so if 0 < |x| < δ, (1/|x|)2 > δ2. so no δ less than 1 will work. but if we choose δ ≥ 1, then (0,1) is a subinterval of (0,δ), and for 0 < |x| < 1, |1/x2| > 1.
since for the particular ε = 1, we can't find a δ, L can't be 0. so L > 0 (since 1/x2) (if L < 0, then on the interval (0,1], 1/x2 would have to take on the value 0 (by the intermediate value theorem) but if x ≠ 0, 1/x2 > 0).
so what happens if δ < 1/(√(2L))?
then 0 < |x| < 1/(√(2L)) means that |1/x2 - L| ≥ |1/x2| - |L|
= 1/|x|2 - L > 1/(1/(√(2L))2) - L = 2L - L = L
now if we choose 0 < ε < L (which we can, since L > 0, for example ε = L/2 would work fine), we have the same problem as before, no δ < 1/(√(2L)) will work, and any larger choice for δ will lead to |x| < 1/(√(2L)) for some x as well (larger δ's don't help).
so there is always some ε > 0 that doesn't have a δ, no matter what we choose for L. so such an L doesn't exist, we cannot define f(0) to be any real number in such a way as to make f(x) = 1/x2 continuous.