Question

Utilizand eventua identitatea: $sina-sinb=2sin\frac{a-b}{2}cos\frac{a+b}{2}$
Sa se calculeze :
$\lim_{x \to \infty} sin\sqrt{x+1} -sin\sqrt{x}$

Answer 1

$\text{Voi folosi o inegalitate cunoscuta:}\\ \\ |\sin a-\sin b| \leq |a-b|\\ \\\\ \Rightarrow\,|\sin \sqrt{x+1}-\sin \sqrt x| \leq |\sqrt{x+1}-\sqrt{x}| \\\\ \Leftrightarrow\,-(\sqrt{x+1}-\sqrt x) \leq \sin \sqrt{x+1}-\sin \sqrt x \leq \sqrt{x+1}-\sqrt x \\ \\ \Leftrightarrow\,\sqrt x - \sqrt{x+1}\leq \sin \sqrt{x+1}-\sin \sqrt x \leq \sqrt{x+1}-\sqrt x\\ \\\\ \sqrt{x+1}-\sqrt{x}<\dfrac{1}{2\sqrt x},\quad \forall x> 0\\ \\ \\\,\Leftrightarrow -\dfrac{1}{2\sqrt x}<\sin \sqrt{x+1}-\sin \sqrt x<\dfrac{1}{2\sqrt x}$

Trecem la limită:

$\Leftrightarrow \,0 < \lim\limits_{x\to \infty}\Big(\sin \sqrt{x+1}-\sin \sqrt x\Big) < 0 \\ \\ \\\Rightarrow \lim\limits_{x\to \infty}\Big(\sin \sqrt{x+1}-\sin \sqrt x\Big) = 0$