Show $\cos(x^2)/(1+ x^2)$ is uniformly continuous on $\Bbb R$.

Question

now here's how I did proceed.

By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$

Then suppose $x, a$ are elements of $\Bbb R. $ Now \begin{align} |f(x) - f(a)| &= \left|\frac{\cos(x^2)}{1 + x^2} - \frac{\cos(a^2)}{1 + a^2}\right| \\&= \left| \frac{\cos(x^2)(1+a^2 )- \cos(a^2)(1+x^2)}{(1 + x^2)(1 + a^2)}\right| \\&≤ \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right| \end{align}

can it be written..? the last step? If not, please help me solve the sum.

Answer 1

Hint: use triangular inequality and $2|x| < 1 + x^2$ to get $|f'(c)| < \dfrac{2}{1+c^2} < 2 \Rightarrow |f(x) - f(a)| < 2|x-a|$ , and conclude.