Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 09/Solution 2

Here is a solution using the limit comparison test, Theorem 1.12.22 in CLP2 . We expect the integrand to behave like $\frac{1}{x}$, whose integral from 1 to $\mathrm{\infty }$ diverges, so we apply the limit comparison test with $f(x)=\frac{x^2+1}{x^3+2}$ and $g(x)=\frac{1}{x}$ (notation as in the theorem statement in CLP). We compute then $$\underset{x\to \mathrm{\infty }}{\lim }\frac{f(x)}{g(x)}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{x^2+1}{x^3+2}}{\frac{1}{x}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2+1}{x^3+2}\cdot x=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^3+x}{x^3+2}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1+1/x^2}{1+2/x^3}=\frac{1+0}{1+0}.$$ Since the above limit exists and is non-zero, the improper integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)\mathrm{d}x$ diverges by the limit comparison test (b).