Question

Nu mai stiu exact cum se rezolvau astfel de probleme. Eu am incercat initial sa amplific cu conjugata, dar nu m-a ajutat prea mult.

Answer 1

$l = \lim\limits_{x\to-\infty}\Big(\sqrt{4x^2+ax+1}-bx\Big)\\ \\ \text{Fac schimbarea de variabila: }x = -t \\ \\ l = \lim\limits_{t\to\infty}\Big(\sqrt{4t^2-at+1}+bt\Big)\\ \\ \sqrt{4t^2-t+1} \approx 2t-\dfrac{1}{4},\quad \text{cand }x\to \infty\\ \\ \text{Fiindca }\Big(2t-\dfrac{1}{4}\Big)^2 = 4t^2-t+\dfrac{1}{8}\approx \sqrt{4t^2-t+1}^2,\quad x \to \infty \\ \\\text{Analog: } \sqrt{4t^2-at+1} \approx 2t-\dfrac{a}{4},\quad \text{cand }x\to \infty \\ \\ \Rightarrow l = \lim\limits_{t\to\infty}\Big(2t-\dfrac{a}{4}+bt\Big)$

$\Rightarrow \lim\limits_{t\to\infty}\Big[t(2+b) - \dfrac{a}{4}\Big] =2 \\ \\ \Rightarrow 2+b = 0\quad \text{si}\quad -\dfrac{a}{4} = 2 \\ \\\Rightarrow \boxed{a = -8}\quad \text{si}\quad \boxed{b = -2}$