epsilon delta limit proof of $\lim_{x \to 0} \arctan{(x)}=0$

Question

Let $\lim x→0 \arctan{(x)}=0$ then there exists $\delta>0$ such that if $0<|x|<\delta$ then $|\arctan{(x)}|< \varepsilon $

I know that,

$|\arctan{(x)}|<\pi/2$

But how do I relate this to $|x|$ so I can find a suitable $\delta$? Because I don't even know how to check if $|\arctan{(x)}|\leq|x|$ is true (for those $x$ near zero).

Answer 1

By deriving $f(x)=\arctan(x)$ with respect to $x$ you get $$f^\prime (x) =\frac{1}{1+x^2} ,$$ so $$0 < f^\prime (x) \le 1$$ and, finally, this means $$|f(x)| = |\arctan(x)| \le |x|.$$ Knowing this, now let $\epsilon >0$, then for all $x$ such that $|x|<\epsilon$, $|\arctan(x)| \le |x| < \epsilon$ just as we wanted; in particular we choose $\delta = \epsilon$.