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.