I don't think you can really prove this.
lim x->0 1/x2 does not exist (DNE) as a finite number;
although we can suspect that it
diverges to infinity.
lim x->0 1/x2 = ∞
really comes from a definition. And many courses say that all limits -> ∞ DNE.
Definition We say that a function f(x) diverges to infinity, denoted by f(x) -> ∞ as x -> a,
if, for every ε > 0, there exists δ > 0 such that |f(x)| > ε whenever 0 < |x - a| < δ.
Consider the function f(x) = 1/x. We suspect that f(x) -> ∞ as x -> 0.
Here a = 0. Let ε > 0 be chosen. Then
|f(x)| = | 1/x | = 1/|x| > ε whenever | x - 0 | = |x| < δ = 1/ε.
Proposition The function f(x) -> ∞ as x -> a if and only if the function 1/f(x) -> 0 as x -> a.
1st attempt - Better Way Below
So using the properties of limits we could also say
lim x->0 1/x2 = (lim x->0 1) / (lim
given lim c = c and lim x->a x = a and lim
x->0 f(x2) = (lim x->0 f(x))2
= 1 / (lim x->0 x)2 = 1 / (0)2 = undefined
But now that I think about it you could use the proposition to show this in a better way.
f(x) -> ∞ if 1/f(x) - > 0
So show that 1/f(x) = x2
Then show that x2 -> 0 (by substitution, or more formally.
Now since 1/f(x) = x2 -> 0 Then f(x) = 1/x2 -> ∞ or DNE