Problem finding the limit of the remainder of a series

Suppose there is a series

$\displaystyle \frac{1}{1+t^2} = 1 -t^2 + t^4 - t^6 + ... + (-1)^n t^{2n} + \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2} $

in which the last term comes from adding the remaining terms as a

geometric series with the first term $\displaystyle a = (-1)^{(n+1)} t^{(2n+2)}$

and ratio $\displaystyle r= -t^2$ Integrating both sides of equation from

$\displaystyle t = 0$ to $\displaystyle t = x$ gives

$\displaystyle tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^n \frac{x^{(2n+1)}}{2n+1} + R_n(x)

$

where

$\displaystyle

R_n(x) = \int_0^x \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}dt

$

The denominator of integrand is greater than or equal to 1;hence

$\displaystyle

\left|R_n(x)\right| \le \int_0^{\left|x\right|} t^{(2n+2)} dt = \frac{\left|x\right|^{(2n+3)}}{2n+3}

$

if $\displaystyle \left|x\right| \le 1$ , the right side of this inequality approaches zero as $\displaystyle n \to \infty$

Therefore

$\displaystyle

\lim _{n \to \infty} R_n(x) = 0\,\, if \left|x\right| \le 1

$

Now my question is how the book found the limit of $\displaystyle R_n(x)$ as 0?

Book showed and I understand that the limit of $\displaystyle \frac{\left|x\right|^{(2n+3)}}{2n+3}$ is 0 when n

approaches $\displaystyle \infty$. But that's only half of it? The author did not calculate

the limit of integral $\displaystyle R_n(x)$ as a whole. I don't understand how he

reached the conclusion? Obviously I'm missing something simple. Is

there any theorem that i'm unaware of? Anyone knows how

he reached the conclusion?