I don't think you can really prove this.
lim _{x->0} 1/x^{2 } does not exist (DNE) as a finite number;
although we can suspect that it
diverges to infinity.
lim _{x->0} 1/x^{2} = ∞
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/x^{2} = (lim_{ x->0} 1) / (lim _{
x->0} x^{2})
given lim c = c and lim _{x->a} x = a and lim
_{x->0} f(x^{2}) = (lim _{x->0} f(x))^{2}
= 1 / (lim _{x->0} x)^{2} = 1 / (0)^{2} = undefined
Better Way
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) = x^{2}
Then show that x^{2} -> 0 (by substitution, or more formally.
Now since 1/f(x) = x^{2} -> 0 Then f(x) = 1/x^{2} -> ∞ or DNE
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