Epsilon-Delta Proof of limit approaching infinity

[RPierre]

**DISCLAIMER - I am super bad at LaTeX**

Homework Statement

Prove

$$\lim_{x \rightarrow \infty}\frac{1}{1+x^2} = 0$$

Homework Equations

I Think I proved it, but I feel like I'm missing something to make this a proof of ALL $$\epsilon$$>0 and not just one case. Maybe I did it right. I really don't know. Just looking for a second opinion and/or advice on $$\epsilon$$-$$\delta$$ proofs.

3.Attempt at a Solution

the definition logically is if $$x > N$$ then $$|f(x) - L| > \epsilon$$ for some $$N,\epsilon > 0$$

Setting $$N=\sqrt{\frac{1-\epsilon}{\epsilon}}$$

x > N

$$\Rightarrow$$ $$x > \sqrt{\frac{1-\epsilon}{\epsilon}}$$

$$\Rightarrow$$ $$x^2 > \frac{1-\epsilon}{\epsilon}$$

$$\Rightarrow$$ $$x^2 + 1 > \frac{1}{\epsilon}$$

$$\Rightarrow$$ $$\frac{1}{1+x^2} < \epsilon$$

$$\frac{1}{1+x^2} = f(x)$$

and since N > 0, and x > N, it is implied x > 0 and therefore $$|f(x)| = f(x)$$

I'm not sure if this is a good enough proof? Thanks in Advance :)

 

[LCKurtz]

Yes, your argument is fine. I would start it and end it slightly differently:

Begin with:

Suppose ε > 0 Then your next line

Let N = ... is OK
Then, if x > N...
...
...
$$\frac{1}{1+x^2} < \epsilon$$

So

$$|\frac{1}{1+x^2}| < \epsilon$$

And stop there.