Question

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Answer 1

$l =\lim\limits_{x\to \infty} x\Big(ax+\sqrt{2+bx+cx^2}\Big) = \\ \\ = \lim\limits_{x\to \infty} x\Bigg(ax+x\sqrt{\dfrac{2}{x^2}+\dfrac{b}{x}+c}}\,\Bigg) =\\ \\ =\lim\limits_{x\to \infty} x^2\Bigg(a+\sqrt{\dfrac{2}{x^2}+\dfrac{b}{x}+c}}\,\Bigg) \\ \\\dfrac{1}{x} =t \Rightarrow t \to 0 \\ \\ l=\lim\limits_{t\to 0}\dfrac{a+\sqrt{2t^2+bt+c}}{t^2} \\ \\ \text{Observam ca }a = -1,\quad c = 1$

$\text{Deoarece trebuie sa avem nedeterminare 0/0 ca sa nu ne dea }+ \infty.\\ \\ l = \lim\limits_{t\to 0}\dfrac{-1+\sqrt{2t^2+bt+1}}{t^2} \overset{\frac{0}{0}}{=} \lim\limits_{t\to 0}\dfrac{\dfrac{4t+b}{2\sqrt{2t^2+bt+1}}}{2t} = \\ \\ = \lim\limits_{t\to 0}\dfrac{4t+b}{4t\sqrt{2t^2+bt+1}} \\ \\ \\b = 0 \text{ deoarece trebuie ca limita sa fie 1}. \\\\ \\ \Rightarrow l = \dfrac{1}{\sqrt{1\cdot 0+0+1}}= 1\\ \\ \Rightarrow a = -1,\quad b = 0,\quad c = 1$