How to show $\int_x ^\infty f(t)dt \to 0$ as $x\to \infty$

Question

If $f\ge 0$ and $f(x)\to 0$ as $x\to \infty$, what is the rigorous way to verify that $\lim_{x\to \infty}\int_x ^\infty f(t)dt=0$ if the function $f$ is not simple to integrate? If the interval of integration was bounded, say $[x,C]$, one could just find a small upper bound for $f(t)$, say, $f(t)\le \varepsilon$ when $t$ is sufficiently large, and then use $\int_x ^Cf(t)dt \le \int _x ^C\varepsilon dt=\varepsilon( C-x)$. The latest expression tends to $0$ as $x\to C$.

When $C=\infty$, one could try to rewrite the original limit as an improper integral $\lim_{x\to \infty} \big(\lim_{C\to \infty}\int_x ^Cf(t)dt\big)\le \lim_{x\to \infty}\big(\lim_{C \to \infty} \varepsilon(C-x)\big)$, but the right hand side looks like an indeterminate form.

Answer 1

If, for example, the integral $\int_0 ^\infty f(t)dt $ is convergent, then we have, for $x>0$:

$\int_x ^\infty f(t)dt=\int_0 ^\infty f(t)dt -\int_0 ^x f(t)dt \to \int_0 ^\infty f(t)dt -\int_0 ^\infty f(t)dt =0$ as $x \to \infty$.

Answer 2

One can underbound initial function with such function, integral of which over $[x, \infty]$ is infinity.