Hey,
I wanted to ask how is it possible to proof that ArcTan(1/x^2) limit when x goes to infinity, is 0 ?
I need a valid proof and not a intuitive proof (Epsilon Delta\ Or limit arithmetic ..)
Thanks a lot
Well, arctangent is the inverse to tangent: y= arctan(1/x^2) is the same as 1/x^2= tan(y). I assume you know that as x goes to infinity, 1/x^2 goes to 0. So what must tan(y) be? And then, what must y be?
Now that's "intuitive" and not a "valid proof" but it is thinking like that that will lead you to a "valid proof". We want to show that in order to make arctan(1/x^2) as close to 0 as we please, $\displaystyle |arctan(1/x^2)|< \epsilon$, which is $\displaystyle -\epsilon< arctan(1/x^2)< \epsilon$, $\displaystyle 0 < 1/x^2< tan(\epsilon)$, $\displaystyle x^2> tan(\epsilon)$, $\displaystyle x> \sqrt{tan(\epsilon)$.
So take $\displaystyle N> \sqrt{tan(\epsilon)}$ and work backwards: if $\displaystyle x> N$ then $\displaystyle |arctan(1/x^2)|< \epsilon$.
You know that $\displaystyle \arctan(x)$ is a continuous function.
And $\displaystyle \arctan(0)=0$ also $\displaystyle \lim _{x \to \infty } \frac{1}{x} = 0$.
Combine these $\displaystyle \left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left[ {\left| x \right| < \delta \to \left| {\arctan (x)} \right| < \varepsilon } \right]$.
And $\displaystyle \left( {\forall \delta > 0} \right)\left( {\exists N \in \mathbb{N}} \right)\left[{x \geqslant N \to \frac{1}{x} < \delta } \right]$
A question for HallsofIvy:
On line 3, You got from 0<1/x^2<tan(e) to x^2>tan(e) .. shouldnt it be x^2 > 1/tan(e) ? And then the final requirement x>1/sqrt(tan(e)) ?