Evaluate $\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$ 
Evaluate $$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$$

My Solution
Denote
$$f(t):=\arctan t.$$
By Lagrange's Mean Value Theorem，we have
$$f\left(\frac{2x^2+5}{x^2+1}\right)-f\left(\frac{2x^2+7}{x^2+2}\right)=f'(\xi)\left(\frac{2x^2+5}{x^2+1}-\frac{2x^2+7}{x^2+2}\right)=\frac{3}{(1+\xi^2)(x^2+1)(x^2+2)}$$
where $$\frac{2x^2+5}{x^2+1}\lessgtr \xi \lessgtr \frac{2x^2+7}{x^2+2}.$$
Here, applying the Squeeze Theorem, it's easy to see$$\lim_{x \to \infty}\xi=2.$$
It follows that
$$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)=\lim_{x \to \infty}\frac{3x^4}{(1+\xi^2)(x^2+1)(x^2+2)}=\frac{3}{5}.$$
Hope to see other solutions.THX.
 A: Recall that by  Arctangent addition formula
$$\arctan\dfrac{2x^2+5}{x^2+1}-\arctan\dfrac{2x^2+7}{x^2+2}=\arctan \left(\frac{\dfrac{2x^2+5}{x^2+1}-\dfrac{2x^2+7}{x^2+2}}{1+\dfrac{2x^2+5}{x^2+1}\dfrac{2x^2+7}{x^2+2}}\right)=$$
$$=\arctan \left(\frac{3}{5x^4+27x^2+37}\right)$$
therefore
$$x^4\left(\arctan\dfrac{2x^2+5}{x^2+1}-\arctan\dfrac{2x^2+7}{x^2+2}\right)=\frac{\arctan \left(\frac{3}{5x^4+27x^2+37}\right)}{\frac{3}{5x^4+27x^2+37}}\frac{3x^4}{5x^4+27x^2+37}\to \frac35$$
A: When $x\to \infty$, 
$$\frac{2x^2+5}{x^2+1}=2+\underbrace{\frac{3}{x^2+1}}_{\to 0},\qquad \frac{2x^2+7}{x^2+2}=2+\underbrace{\frac{3}{x^2+2}}_{\to0},$$
and the Taylor approximation of $\arctan(2+t)$ around $0$ is given by
\begin{align*}
\arctan(2+t)=\arctan 2+(\arctan(t+2))'|_{t=0}t+o(t)=\arctan 2+\frac t5+o(t).
\end{align*}
Set $t=\dfrac{3}{x^2+1}$ and $t=\dfrac{3}{x^2+2}$, we have $o(t)=o(x^{-2})$ and
\begin{align*}
\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}&=\frac 35\underbrace{\left(\frac{1}{x^2+1}-\frac{1}{x^2+2}\right)}_{:=g(x)}+o(x^{-2}).\\
\end{align*}
Therefore,
\begin{align*}
L=\frac35\underbrace{\lim_{x\to\infty}x^4g(x)}_{=1}+\underbrace{\lim_{x\to\infty}\frac{o(x^{-2})}{x^{-4}}}_{=\lim\limits_{t\to0}\frac{o(t)}{t^2}=0}=\frac35.
\end{align*}
