Prove that $$x - \frac{x^3}3 < \arctan x < x$$ for every $x>0$?
I tried taking the limit of $x-x^3/3$ and $\arctan(x)$ as $x$ approaches $0$, but I get $0$ which makes sense since they're both $0$ at $x=0$
I'm not sure what else to do algebraically. Would appreciate some help.
Note that $$1-t^2 \le\frac{1}{1+t^2} \le 1$$ for a real $t$ (the inequalities hold with equality only when $t=0$). Therefore, $$\int_0^{x}(1-t^2)\, dt \le \int_0^{x}\frac{1}{1+t^2}\, dt \le \int_0^{x}1\, dt$$ for $x \ge 0$. That is, $$x-\frac{x^3}{3} \le \arctan(x) \le x.$$
The inequalities hold with equality only when $x=0$.
As suggested in the comments, for $x > 0$, we have $$\int_{x/2}^{x}(1-t^2)\, dt < \int_{x/2}^{x}\frac{1}{1+t^2}\, dt < \int_{x/2}^{x} 1\, dt.$$ Therefore, $$\int_{0}^{x}(1-t^2)\, dt < \int_{0}^{x}\frac{1}{1+t^2}\, dt < \int_{0}^{x} 1\, dt$$ for $x > 0$.
Hint:
$$\frac{d}{dx}\arctan x=(1+x^2)^{-1}$$
$$(1+x^k)^n=\sum_{r=0}^\infty\dfrac{n(n-1)\cdots(n-r+1)}{r!}x^{kr}$$
(1) Let $$f(x)=\arctan x,~~~x>0.$$
By Lagrange's mean value theorem，we obtain $$f(x)-f(0)=f'(\xi_1)(x-0),$$where $0<\xi_1<x.$ Thus, $$\arctan x=\frac{1}{1+\xi_1^2}\cdot x<x.$$
(2) Let $$g(x)=\arctan x-x+\frac{x^3}{3},~~~x>0.$$
By Lagrange's mean value theorem，we obtain $$g(x)-g(0)=g'(\xi_2)(x-0),$$where $0<\xi_2<x.$ Thus, $$\arctan x-x+\frac{x^3}{3}=\frac{\xi_2^4}{\xi_2^2+1}\cdot x>0,$$namely,$$\arctan x>x-\frac{x^3}{3}.$$
Combining (1) and (2), we are done.
