Comparison Test for Improper Integrals — Question 2

Question 2 – Solution

Step 1: Choose a comparison. The dominant powers give $$\frac{3x^2}{x^4}=\frac3{x^2},$$ so use $g(x)=1/x^2$.

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