<s>[INST] Showing $f(x) = \frac{1}{1 + x^2}$ is continuous I need to show with a $\epsilon$-$\delta$ proof that the function $$f : \mathbb{R} \rightarrow \mathbb{R} : x \mapsto \frac{1}{1 + x^2} $$ is continuous. 
Here is what I did. Let $a \in \mathbb{R}$ be arbitrary. Then for every $x \in \mathbb{R}$ we have $$ | f(x) - f(a) | = \bigg| \frac{1}{1 + x^2} - \frac{1}{1 + a^2} \bigg| = \frac{| a^2 - x^2 |}{(1+x^2)(1+a^2)} \leq \frac{| x - a| (|a| + |x|)}{(1+x^2)(1+a^2)}. $$ Now I think I need to use the trick that $$ \frac{|t|}{1 + t^2} \leq \frac{1}{2} $$ for all $t \in \mathbb{R}$. If I do that here, I still get something complicated: $$ ... = \frac{|x - a| |a|}{ (1+x^2)(1+a^2)} + \frac{|x-a| |x|}{(1+x^2)(1+a^2)} \leq \frac{|x-a|}{ 2 (1+x^2)} + \frac{|x-a|}{2(1+a^2)}. $$ I still got my RHS in terms of $x$, while I only want it in terms of $|x-a|$ and $a$ so than I can pick my $\delta$.
Any help please?
 [/INST] You can use 
$$\frac{|x-a|}{ 2 (1+x^2)} + \frac{|x-a|}{2(1+a^2)}\le \frac{|x-a|}{2} + \frac{|x-a|}{2}. $$
Or use 
$$\frac{| x - a| (|a| + |x|)}{(1+x^2)(1+a^2)}\le \frac{2|x-a| \max\{|a|,|x|\}}{(1+a^2)(1+x^2)} \le 2|x-a|$$
and then choose $\delta = \epsilon/2$. 
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