Comparison Test for Improper Integrals — Question 7

Question 7 – Solution

Step 1: Choose a comparison. The dominant powers give $$\frac{x^3}{x^5}=\frac1{x^2},$$ so let $g(x)=1/x^2$.

Step 2: Compute the limit ratio. $$\begin{align*} L&=\lim_{x\to\infty} \frac{(x^3+4)/(x^5+x)}{1/x^2}\\ &=\lim_{x\to\infty}\frac{x^5+4x^2}{x^5+x}\\ &=\lim_{x\to\infty} \frac{1+4/x^3}{1+1/x^4}=1. \end{align*}$$ Step 3: Apply limit comparison. Since $\int_1^\infty x^{-2}\,dx$ converges and $0<L<\infty$, the original integral converges. $$\boxed{\text{converges}}$$