Question

Fie f:[0,∞) -> [0,∞] o functie cu proprietate ca f(x^2+x) ≤ x, ∀x≥0
Daca: L = $\lim_{x \to \ 0} \frac{f(x)}{\sqrt{x} } \\x\ \textgreater \ 0$
Cat este L ?

Answer 1

$f:[0,+\infty)\to [0,+\infty),\,\, f(x^2+x)\leq x,\quad \forall x\geq 0\\ \\ \lim\limits_{x\searrow 0}\dfrac{f(x)}{\sqrt x}=? \\ \\ \\ \dfrac{f(x^2+x)}{\sqrt{x^2+x}} \leq \dfrac{x}{\sqrt{x^2+x}}\\ \\ x^2+x = t \Rightarrow x^2 = t-x\Rightarrow x = \sqrt{t-x}\Rightarrow\\ \\ \Rightarrow x = \sqrt{t-\sqrt{t-\sqrt{t-\sqrt{t-....}}}}$

$\Rightarrow 0<\dfrac{f(x)}{\sqrt{x}}\leq \dfrac{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-...}}}}}{\sqrt{x}}\\ \\ 0 <\lim\limits_{x\searrow 0}\dfrac{f(x)}{\sqrt x}\leq \lim\limits_{x\searrow 0}\dfrac{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-...}}}}}{\sqrt{x}} \\ \\ l_1 =\lim\limits_{x\searrow 0}\dfrac{\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-...}}}}}{\sqrt{x}}= \\ \\ = \sqrt{ \lim\limits_{x\searrow 0}\dfrac{x-\sqrt{x-\sqrt{x-\sqrt{x-...}}}}{x}}$

Suntem în cazul 0/0.

Aplicăm L'Hôpital.

$y = \sqrt{x-\sqrt{x-\sqrt{x-....}}} \\ y = \sqrt{x-y}\\ y^2 = x-y\\ 2yy'=1-y' \\ 2yy'+y' = 1 \\ y'(2y+1) = 1 \\ y' = \dfrac{1}{2y+1}\\ \\ \Rightarrow l_1 = \sqrt{ \lim\limits_{x\searrow 0}\dfrac{1-\dfrac{1}{2\sqrt{x-\sqrt{x-\sqrt{x-...}}}+1}}{1}} =\sqrt{\dfrac{1-\dfrac{1}{2\cdot 0+1}}{1}} = 0 \\ \\ \\\Rightarrow 0<\lim\limits_{x\searrow 0}\dfrac{f(x)}{\sqrt x}\leq 0 \\ \text{}\quad\quad\,\,\,\searrow\quad\quad\quad\swarrow\\ \\ \text{}\quad\quad\quad\,\,\,\, \boxed{L = 0}$