Another $\varepsilon$-$\delta$ argument for $h(x)=\frac{1}{1+x^2}$ Given $h(x)=\frac{1}{1+x^2}$, prove that $h:\mathbb{R}\rightarrow\mathbb{R}$ satisifies the $\varepsilon$-$\delta$ criterion on $\mathbb{R}$.
So I have a rather difficult time with these $\varepsilon$-$\delta$ arguments. I understand the basic understanding but finishing out the arguments continues to elude me. This is what I have so far.
Let $\epsilon>0$ be given and let $x_0$ be an arbitrary point in $\mathbb{R}$. Consider all $x\in\mathbb{R}$ such that $|x-x_0|<\delta$ for some $\delta>0$. Then
$$|f(x)-f(x_0)|= \left|\frac{1}{1+x^2}-\frac{1}{1+x_0^2}\right|=\left|\frac{x^2-x_0^2}{(1+x^2)(1+x_0^2)}\right|=\frac{|x-x_0||x+x_0|}{|1+x^2||1+x_0^2|}.$$
First I realized that I skipped some of the algebra. I did so simply to save space. Now I have $\delta$ control over the $|x-x_0|$ term in the numberator and since $x\in(x_0-\delta,x_0+\delta)$ I should be able to bound the other three terms but I have a rather difficult time working out what those bounds and and stating things precisely.
 A: $$|f(x)-f(x_0)|=|\frac{1}{1+x^2}-\frac{1}{1+x_0^2}|=|\frac{x^2-x_0^2}{(1+x^2)(1+x_0^2)}|=\frac{|x-x_0||x+x_0|}{|1+x^2||1+x_0^2|}=$$
$$|x-x_0|\bigg (\frac{|x+x_0|}{|1+x^2||1+x_0^2|}\bigg)$$
Knowing that, $$|x-x_0|<\delta$$ try to put a limit on  $$ \frac{|x+x_0|}{|1+x^2||1+x_0^2|}\le M$$
Then pick your $\delta <\epsilon /M$
A: Let $\epsilon>0$. We want to find such $\delta>0$ such that for all $x\in \mathbb R$ such that $|x-x_0|<\delta,$ we have $$\left|\frac{x^2-x_0^2}{(1+x^2)(1+x_0^2)}\right|<\epsilon.$$
I claim that $\delta=\sqrt{\epsilon+x_0^2}-|x_0| > 0$ works. The denominator in the above expression is at least one, so let's focus on the numerator: $|x^2-x_0^2|=|x-x_0|\cdot |x_0+x|$. The first term should be bounded by the $\delta$ we want to find. The second one is bounded by $2|x_0|+\delta,$ so we want to find a $\delta>0$ such that $\delta\cdot (2|x_0|+\delta)\le\epsilon$.
A: Correct me if wrong:
$|f(x)-f(x_0)|= \dfrac{|x-x_0||x+x_0|}{|1+x^2||1+x_0^2|} $
$\le |x-x_0||x+x_0| =$
$|x-x_0||(x-x_0)+2x_0|\le$
$|x_-x_0|(|x-x_0| +|2x_0|).$
Choose $\delta \lt \min (1, \dfrac{\epsilon}{1+2|x_0|})$
Then $|x-x_0| \lt \delta$ implies
$|f(x)-f(x_0)| \le$
$ |x-x_0|(|x-x_0| +2|x_0|) \lt$
$\delta (1+2|x_0|) \lt \epsilon.$
