THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) The function $f:(0,\infty )\to \mathbb{R}$, $f(x)=2x+2x\cos x^2$, is clearly continuous, $$\underset{x\to \infty }{\lim }\frac{1}{x^2}{\int }_0^{x}f(t)dt=\underset{x\to \infty }{\lim }\left(1+\frac{\sin x^2}{x^2}\right)=1,$$ but $f(x)/x$ has obviously not a limit as $x\to \infty$. Another example is the function $x\mapsto 4x(\cos x{)}^2$, $x>0$; verifications are routine, and hence omitted.

b) Let $g:(0,\infty )\to \mathbb{R}$, $g(x)=x^{-2}{\int }_0^{x}f(t)dt-1$, so $g(x)\to 0$ as $x\to \infty$. Fix a small enough $\epsilon >0$. Since $g(x)\to 0$ as $x\to \infty$, if $x$ is large enough, then $$|g(x-\epsilon x)|<{\epsilon }^2,|g(x)|<{\epsilon }^2\text{and}|g(x+\epsilon x)|<{\epsilon }^2.(\ast )$$ Consider such an $x$; since $f$ is increasing, $$\frac{1}{\epsilon x}{\int }_{x-\epsilon x}^{x}f(t)dt\le f(x)\le \frac{1}{\epsilon x}{\int }_x^{x+\epsilon x}f(t)dt.$$ Alternatively, but equivalently, in terms of $g$, $$2-\epsilon +\frac{g(x)}{\epsilon }-(1-\epsilon {)}^2\cdot \frac{g(x-\epsilon x)}{\epsilon }\le \frac{f(x)}{x}\le 2+\epsilon +(1+\epsilon {)}^2\cdot \frac{g(x+\epsilon x)}{\epsilon }-\frac{g(x)}{\epsilon },$$ so, by (*), $$2-2\epsilon -\epsilon (1-\epsilon {)}^2<f(x)/x<2+2\epsilon +\epsilon (1+\epsilon {)}^2.$$