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Concursul 'Marian Tarina' 2008 pb 4
Posted: Mon May 19, 2008 1:34 pm
by Radu Titiu
Fie \( f: (0,\infty) \to [0,\infty) \) o functie marginita. Daca
\( \lim_{x \to 0} \left( f(x) -\frac{1}{2}\sqrt{f \left( \frac{x}{2}\right)}\right)=0 \)
\( \lim_{x \to 0} \left( f(x) -2f^2(2x)\right)=0 \),
atunci \( \lim_{x \to 0}f(x)=0 \).
Posted: Mon May 19, 2008 3:58 pm
by Bogdan Cebere
Notam \( \frac{x}{2} \) cu t si aducem 2 la numarator, adica \( f(x) -\frac{1}{2}\sqrt{f \left( \frac{x}{2}\right)}= 2 f(2t) -\sqrt{f ( {t})} \). Inmultim relatia cu \( 2 f(2t) +\sqrt{f ( {t})} \) si facem observatiile:
* \( f \) nu poate tinde la infinit deoarece ar contrazice faptul ca e marginita.
* daca \( 2 f(2t) +\sqrt{f ( {t})}=0 \), prin trecere la limita obtinem ca \( \lim_{t \to 0}f \left( t)=0 \), deoarece functia este pozitiva.
* daca \( 2 f(2t) + \sqrt{f ( {t})} \neq 0 \), atunci prin inmultirea cu \( 2 f(2t) -\sqrt{f ({t})} \) obtinem ca \( \lim_{t \to 0} \left(4 f^2(2t) -f(t)\right)=0 \). Combinand cu relatia 2 din ipoteza obtinem ca \( \lim_{t \to 0}f(t)=0 \).