Petrovcic2x

2022-06-22

Can use FToC to evaluate $\underset{x\to \mathrm{\infty}}{lim}\frac{{\int}_{0}^{x}\phantom{\rule{mediummathspace}{0ex}}f\left(t\right)dt}{{x}^{2}}$?

Jayce Bates

Beginner2022-06-23Added 18 answers

Because $f(x)\ge \frac{1}{2}{x}^{2}-2$, $\underset{x\to \mathrm{\infty}}{lim}f(x)=\mathrm{\infty}$. Hence the limit is of the indeterminate form $\frac{\mathrm{\infty}}{\mathrm{\infty}}$ and we apply L'Hopital's rule. We find the derivative of the numerator by the FTC, getting

$\underset{x\to \mathrm{\infty}}{lim}\frac{{\int}_{0}^{x}f(t)dt}{{x}^{2}}=\underset{x\to \mathrm{\infty}}{lim}\frac{f(x)}{2x}=\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{cos}x}{2x}-\frac{1}{2x}+\frac{x}{4}$

These three limits evaluate to $0$ (by the squeeze theorem), $0$, and $+\mathrm{\infty}$, respectively; hence their sum has limit $+\mathrm{\infty}$.

$\underset{x\to \mathrm{\infty}}{lim}\frac{{\int}_{0}^{x}f(t)dt}{{x}^{2}}=\underset{x\to \mathrm{\infty}}{lim}\frac{f(x)}{2x}=\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{cos}x}{2x}-\frac{1}{2x}+\frac{x}{4}$

These three limits evaluate to $0$ (by the squeeze theorem), $0$, and $+\mathrm{\infty}$, respectively; hence their sum has limit $+\mathrm{\infty}$.