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Prove: If g is strictly positive then $\int_{2}^{x}o(g)\,dt = o\left(\int_{2}^{x}g\,dt\right)$.

I understand this question to mean: prove that $f=o(g)$ implies $F = o(G)$ and conversely. Please advise if this is not correct. If it is, then

Suppose $\lim\limits_{x \to \infty}{\dfrac{f}{g}} = 0 $ as $ x \to \infty $. Since $g \gt 0$, then $\lim\limits_{x \to \infty}{G} = \infty.$

These conditions allow application of l'Hôpital, as given in Rudin (3rd), p. 113. We can conclude that

$\lim_{x \to \infty} \frac{F}{G} = 0.$

Now suppose that $F=o(G)$. Again we have that $g \gt 0$ and so $\lim_{x\to\infty}{G} = \infty$.

If we can say that $F$ is at least piecewise continuous and differentiable then I suppose l'Hôpital applies and $f = o(G)$. But I can't state precisely why $F=o(G)$ and $ G \to \infty $ give that $F$ satisfies the requirements for application of L'Hôpital.

For example, why might we conclude that $\lim\limits_{x \to \infty} F = \infty$?

Thanks for any suggestions.

This problem is from a comment of Gerry Myerson. Please do not award me points for it.

G.M.'s original comment here

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    For x > x_e, F'/G' = [eG(x+h)-eG(x)]/[G(x+h)-G(x)], etc. Thank you.2011-11-21

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