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Fie \( f:\mathbb{R}\to\mathbb{R}_+ \) o functie continua, \( l>0 \) si \( n\in\mathbb{N} \). Fie \( F \) primitiva functiei \( f^n \) cu proprietatea ca \( F(0)=0 \). Sa se arate ca daca \( \lim_{x\to\infty}f(x)F(x)=l \), atunci avem \( \lim_{x\to\infty}xf^{n+1}(x)=\frac{l}{n+1} \).

Concurs Gh. Dumitrescu 2004

We have the following
\( F^{\prime}F^{n}\sim l^{n}>0 \)
\( F^{n+1}(x)\sim (n+1)l^{n}x \)
\( F(x)\sim (n+1)^{\frac{1}{n+1}}l^{\frac{n}{n+1}}x^{\frac{1}{n+1}} \)

Using that
\( f(x)F(x)\sim l \)
we have
\( f(x)\sim \frac{1}{(n+1)^{\frac{1}{n+1}}}l^{\frac{1}{n+1}}\frac{1}{x^{\frac{1}{n+1}}} \)

Taking the n+1 power
\( x{f(x)}^{n+1}\sim \frac{l}{n+1} \).