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 $\infty$ 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

$$\lim _{x\to \infty }{\frac {f(x)}{g(x)}}=\lim _{x\to \infty }{\frac {\frac {x^{2}+1}{x^{3}+2}}{\frac {1}{x}}}=\lim _{x\to \infty }{\frac {x^{2}+1}{x^{3}+2}}\cdot x=\lim _{x\to \infty }{\frac {x^{3}+x}{x^{3}+2}}=\lim _{x\to \infty }{\frac {1+1/x^{2}}{1+2/x^{3}}}={\frac {1+0}{1+0}}.$$

Since the above limit exists and is non-zero, the improper integral $\int _{1}^{\infty }f(x)\mathrm {d} x$ diverges by the limit comparison test (b).