why the integral $\int_{n=0}^{\infty} \frac{dx}{x^8 + \sqrt{x}}$ is converge?

Question

How to show that this integral $\int_{n=0}^{\infty} \frac{dx}{x^8 + \sqrt{x}}$ is divergent? I try using $g(x)=\frac{1}{x^8}$ and then $\lim_{x \to {\infty}} \frac{\frac{dx}{x^8 + \sqrt{x}}}{\frac{1}{x^8}}$ = L (L-constant), but $\int_{n=0}^{\infty} \frac{dx}{x^8}$ is divergent, so my answer is diverge, but the correct one is converge,

Why?

Answer 1

For $x\geq 1$, you should use $$ \frac1{x^8+\sqrt{x}}\leq \frac1{x^8} $$ and for $0<x<1$, you should use $$ \frac1{x^8+\sqrt{x}}\leq \frac1{\sqrt{x}}. $$ Hence, $$ \int_0^\infty\frac1{x^8+\sqrt{x}}~dx \leq \int_0^1\frac1{\sqrt{x}}~dx+\int_1^\infty\frac1{x^8}~dx <\infty. $$

Answer 2

Close to $x=0$, you have by Taylor $$\frac{1}{x^8+\sqrt{x}}=\frac{1}{\sqrt{x}}-x^7+O\left(x^{10}\right)$$ So, no problem.