1. ## Trigonometry and continuity

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

Stuck on this one

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

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

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

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

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

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