$g(x) = 1/(1+x^2)$ is continuous everywhere epsilon delta approach I have a function 
g: R→R given by the function $g(x) = 1/(1+x^2)$.
I want to prove that this is continuous everywhere. I was reading my real analysis textbook and it seems like a great approach would be to do the delta-epsilon method. I don't fully understand how to use this method. 
I know that I can choose an ε > 0 and then I need to find a corresponding δ. How would I show that at a point $p$, $|p - x| < δ$ then $|g(p) - g(x)| < ε$. 
Thanks for the help!   
 A: Hint: $$|g(p)-g(x)|=\frac{|x^2-p^2|}{(1+p^2)(1+x^2)} \leq |x^2-p^2|$$
A: 
Show that $g(x)=\frac{1}{1+x^2}$ is continuous everywhere using the $\delta$-$\epsilon$ definition.

We can compute $|g(p)-g(x)|$ to be
$$
|g(p)-g(x)|
  =
\left|
\frac{1}{1+p^2} - \frac{1}{1+x^2}
\right|
  =
\frac{|x^2-p^2|}{(1+p^2)(1+x^2)}
  =
\frac{|x-p||x+p|}{(1+p^2)(1+x^2)}
$$
By the triangle inequality, $|a+b|\leq|a|+|b|$. So,
$$
|g(p)-g(x)|
  \leq
\frac{|x-p|(|x|+|p|)}{(1+p^2)(1+x^2)}
  =
|x-p|\left(
\frac{|x|}{(1+p^2)(1+x^2)} + \frac{|p|}{(1+p^2)(1+x^2)}
\right)
$$
And since $a<1+a^2$, we have
$$
|g(p)-g(x)|
  <
|x-p|\left(
\frac{1}{(1+p^2)} + \frac{1}{(1+x^2)}
\right)
$$
There's one more simplification to be made, and then your choice of $\delta$ is obvious for which $|p-x|<\delta$ implies $|g(p)-g(x)|<\epsilon$ will be obvious.
A: Let a $\in \mathbb{R}$ and $\varepsilon > 0$. Set $\delta := \min\{1, \frac{\varepsilon}{ 2 |a| +1}\}$.
$$\forall x \in \mathbb{R}: |x-a|<\delta \implies |f(x)-f(a)|=|\frac{1}{1+x^2}-\frac{1}{1+a^2}|=\frac{|a^2-x^2|}{(1+x^2)(1+a^2)}<|a^2-x^2|\\=|x-a||x+a|< \delta |x-a+2a| \leq \delta (|x-a|+2|a|)\leq \delta(1+2|a|)\leq \frac{\varepsilon}{2|a|+1}(1+2|a|)=\varepsilon.$$
The last two inequalities are valid because $\delta\leq 1$, as well as $\delta \leq \frac{\varepsilon}{ 2 |a| +1}$, since $\delta$ is defined as their minimum.
