Limit of a function

Sep 2009
I need help with the last half of the proof.

Prove that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}\) does not exist.

Assume to the contrary that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}\) this exists. Then there should exists a real number \(\displaystyle L\) such that \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}=L\). Let \(\displaystyle \epsilon =1\). Then there exist \(\displaystyle \delta >0\). If \(\displaystyle x\) is a real number for which \(\displaystyle 0<|x|<\delta\), then \(\displaystyle |\frac{1}{x^2}-L|<\epsilon=1\). Choose an integer \(\displaystyle n >\lceil 1/\delta \rceil \geq 1\). Since \(\displaystyle n>1/\delta\), it follows that \(\displaystyle 0<1/n<\delta\).

Let \(\displaystyle x=1/n\). I want to generate contradiction by showing that

\(\displaystyle |\frac{1}{x^2}-L|=|n^2-L|>1\)

With \(\displaystyle n>1\) and \(\displaystyle L\geq 0\), how can show that \(\displaystyle |n-L| >1\)
Sep 2009
Using the definition you provided:
\(\displaystyle \displaystyle{{(\forall L>0)(\exists\delta>0)\left(0<x<\delta\Rightarrow \frac{1}{x^2}=n>L\right)}.\)
We know that no matter how large \(\displaystyle L\) is, there will always be an integer \(\displaystyle n\) such that \(\displaystyle n>L\), but how do we show it?

In regard to \(\displaystyle L\), as we see in the graph, the range \(\displaystyle f(x) \subset \mathbb{R}^+\). Is it necessary that \(\displaystyle L\) be positive? The reason I ask is that if \(\displaystyle L\) need not be positive, then the proof is complete.


MHF Hall of Honor
Jun 2010
You really don't need the n in there.

Go like this:

Let L>0. Let delta = 1 / sqrt(L). Show that if 0 < x < delta, the desired inequality holds.
  • Like
Reactions: novice