Mathematics

The key is limit comparison: rewrite the integrand by factoring out the dominant term in both numerator and denominator, then compare it to a simpler power function whose convergence you already know. Since $x^{3/2}$ dominates $x$ in the denominator for large $x$, the integrand behaves like $\frac{\sqrt{x}}{x^{3/2}} = \frac{1}{x}$ asymptotically—and you can verify this rigorously using the limit comparison test with $g(x) = \frac{1}{x}$.

The key is limit comparison: rewrite the integrand by factoring out the dominant term in both numerator and denominator, then compare it to a simpler power function whose convergence you already know. Since $x^{3/2}$ dominates $x$ in the denominator for large $x$, the integrand behaves like $\frac{\sqrt{x}}{x^{3/2}} = \frac{1}{x}$ asymptotically—and you can verify this rigorously using the limit comparison test with $g(x) = \frac{1}{x}$.