1. ## Trigonometry and continuity

Prove $f(x)=\frac{sin x}{1+x^2}$ is uniformly continuous over R

Stuck on this one

So I'm trying to show for appropiate $\delta$, and $|x-a|<\delta$, that $|f(x)-f(a)|<\epsilon$

So
$|\frac{sin x}{1+x^2}-\frac{sin a}{1+a^2}|
$

$=|\frac{sinx-sina+a^2sinx-x^2sina}{(1+x^2)(1+a^2)}|
$

Known that $|x-a|\geq{|sinx-sina|}$

Using triangle inequality and considering the numerator only
$|sinx-sina+a^2sinx-x^2sina|\leq{|x-a|+ |a^2sinx-x^2sina|}
$

And here I'm stuck.
2. You could instead try to show that the derivative $f'(x)$ is bounded. If this is the case, then by the Mean Value theorem, $|f(x)-f(y)|=|f'(c)||x-y|\leq M|x-y|$. Choosing $\delta=\varepsilon/M$ gives $|f(x)-f(y)|\leq M|x-y|, and we have uniform convergence.
Try to show that $f'(x)$ is bounded.