# asymptotic behavior of function

I'm preparing for my calculus exam. And I need prove this theorem.

Theorem:

Suppose that $f(x)$ is monotonic function at $x\in(a;\infty)$, integral $\int_{a}^{\infty}f(x)dx$ is converge. Prove that $f(x)=o\left(\frac{1}{x}\right)$ when $x\rightarrow\infty$

(here, we used "O" notation)

I tried to use the Cauchy criterion, but I did not come to anything. I would be happy if you show me the way to solve this problem.

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Sketch:

Assume without loss of generality that $f \ge 0$, and $a=1$.

Then, because $f(x)$ is monotonic (decreasing), $I=\int_1^\infty f(x) dx \ge f(2) + 2 f(4) + 4 f(8) + \cdots = \frac{1}{2}\sum_{n=1}^{\infty} 2^n f(2^n)$

Because this series converges, $2^n f(2^n) \to 0$. And, again, because $f(x)$ is monotonic, this implies $x f(x) \to 0$, or $f(x) = o(1/x)$

(This is very similar to did's answer)

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Hint: We have $\int_a^\infty fdx<\int_a^\infty \frac{dx}{x}$, so look at the Riemann sums to show that eventually (i.e. for all $x$ greater than some $n\in\mathbb{R}$) $f(x)<\frac{1}{x}$

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$\frac1{2x}<\frac1x$ but it does not mean that $\frac{1}{2x} = o(\frac 1x)$ –  Ilya Jan 9 '13 at 16:54
Assume without loss of generality that $f\geqslant0$, hence $f$ is nonincreasing.
Then $\displaystyle\int_x^{2x}f(t)\mathrm dt\leqslant\int_x^{+\infty}f(t)\mathrm dt$ for every $x\geqslant a$ and $\displaystyle\int_x^{+\infty}f(t)\mathrm dt\to0$ hence $\displaystyle\int_x^{2x}f(t)\mathrm dt\to0$. For every $x$, the sense of variation of $f$ indicates that $\displaystyle\int_x^{2x}f(t)\mathrm dt\geqslant\int_x^{2x}f(2x)\mathrm dt=xf(2x)$ hence $xf(2x)\to0$. This implies that $xf(x)\to0$.