Thread: Limit of a function

1. Limit of a function

I need help with the last half of the proof.

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

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

Let $x=1/n$. I want to generate contradiction by showing that

$|\frac{1}{x^2}-L|=|n^2-L|>1$

With $n>1$ and $L\geq 0$, how can show that $|n-L| >1$

2. See your other thread. The proof will be the same, essentially.

3. Using the definition you provided:

$\displaystyle{{(\forall L>0)(\exists\delta>0)\left(0L\right)}.$
We know that no matter how large $L$ is, there will always be an integer $n$ such that $n>L$, but how do we show it?

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

4. 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.