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...
...
...
<br /> \frac{1}{1+x^2} &lt; \epsilon <br />

So

<br /> |\frac{1}{1+x^2}| &lt; \epsilon <br />

And stop there.