Help with the limit of an integral

Question

I am trying to evaluate this limit
$$\lim \limits_{x \to \infty} \frac {1}{\ln x} \int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt=? $$
Any help will be appreciated.

Answer 1

Let $f(x)=\int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt$. It is easy to conclude $\lim_{x\to\infty} f(x)=\infty$. Now use L'Hopitals rule: $$\lim \limits_{x \to \infty} \frac {1}{\ln x} \int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt=\lim \limits_{x \to \infty}\frac{2x\cdot \frac{x^{10}-x^4+8}{2x^{12}+x^4+4}}{\frac{1}{x}}=\lim \limits_{x \to \infty} \frac{2x^{12}-2x^6+16x^2}{2x^{12}+x^4+4}=\frac{2}{2}=1$$

Answer 2

Hint: Compare with the integral of just $t^5/(2 t^6)$ as $t$ goes to infinity. Show that the rest really doesn't matter.