# Limit of a function

#### novice

I need help with the last half of the proof.

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

Proof:
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 $$\displaystyle |n-L| >1$$

#### Ackbeet

MHF Hall of Honor

#### novice

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.

#### Ackbeet

MHF Hall of Honor
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.

• novice