Limit of a function

**novice** · July 22nd 2010, 10:57 AM

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 $|n-L| >1$

**Ackbeet** · July 22nd 2010, 11:53 AM

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

**novice** · July 22nd 2010, 12:30 PM

Using the definition you provided:

$\displaystyle{{(\forall L>0)(\exists\delta>0)\left(0<x<\delta\Rightarrow \frac{1}{x^2}=n>L\right)}.$

We know that no matter how large $L$ 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.

**Ackbeet** · July 22nd 2010, 12:36 PM

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.

Thread: Limit of a function

LinkBack

Thread Tools

Display

Limit of a function

Similar Math Help Forum Discussions

Absolute value of the limit of a function is equal to the absolute value of the limit

Finding limit of this function, using Limit rules

limit of function of a sequence equals the function of the limit of that sequence?

[SOLVED] limit of function exists, but limit of its derivative doesn't.

Limit, Limit Superior, and Limit Inferior of a function

Search Tags