Contradictory results in testing convergence of improper integral

Question

I was testing the convergence of the integral $$\int_0^{+\infty}\frac{\sqrt{x+1}}{1+2\sqrt x+x^2}dx$$ by calculating the limit $$\lim_{x\to+\infty}\frac{\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}}{\frac{1}{x^{3/2}}}=1>0$$ and by noting that $$\int_0^{+\infty}\frac{1}{x^{3/2}}=+\infty$$ therefore the original integral has to be divergent as well. However Mathematica seems to evaluate the original integral to 2.04517. What is wrong here?

Answer 1

The limit you have found $$ \lim_{x\to+\infty}\frac{\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}}{\frac{1}{x^{3/2}}}=1 $$ implies that, as $x \to \infty$, $$ \frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}\sim{\frac{1}{x^{3/2}}} $$

you can't deduce it holds as $x \to 0$. Therefore, we have that, for any large fixed $B>0$, $$ \int_B^\infty\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}\:dx, \quad\int_B^\infty {\frac{1}{x^{3/2}}}\:dx, $$ are both convergent. Observe that the integrand $\dfrac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}$ is a continuous function as $x \to 0^+$, giving the convergence of the initial integral over $(0,\infty)$.