Question

daca se poate sa ma ajutati la punctul 1 sub 3​

Answer 1

a)

$y = mx+n$

$f(x) = \sqrt{x^2+x+1}-x \\\\ \\ \Big(x-\dfrac{1}{2}\Big)^2 = x^2-x+\dfrac{1}{4} \approx x^2-x+1,\quad \text{cand }x\to +\infty \\ \\ \Rightarrow x-\dfrac{1}{2} \approx \sqrt{x^2-x+1},\quad \text{cand }x\to +\infty \\ \\\\ \lim\limits_{x\to -\infty} \Big[\sqrt{x^2+x+1}-x-(mx+n)\Big]=0\\ \\ x = -t \Rightarrow t\to +\infty\\ \\ \lim\limits_{t\to +\infty}\Big(\sqrt{t^2-t+1}+t+mt-n\Big) =0$

$\lim\limits_{t\to +\infty}\Big(t-\dfrac{1}{2}+t+mt-n\Big) =0\\ \\ \lim\limits_{t\to +\infty}\Big[(2+m)t-\Big(\dfrac{1}{2}+n\Big)\Big] =0\\ \\\\ \Rightarrow 2+m = 0\quad \text{si}\quad \dfrac{1}{2}+n = 0 \\ \Rightarrow m = -2\quad \text{si}\quad n = -\dfrac{1}{2} \\ \\ \Rightarrow m = -2x-\dfrac{1}{2},\quad \text{asimptota oblica spre }-\infty$

b)

$(\sqrt{x^2+x+1})^2 = x^2+x+1 \\ \\ \Big(x+\dfrac{1}{2}\Big)^2 = x^2+x+\dfrac{1}{4} \\ \\ \\x^2+x+1 > x^2+x+\dfrac{1}{4}\Bigg|\sqrt{(\,)} \\\\ \Rightarrow \sqrt{x^2+x+1} > \sqrt{x^2+x+\dfrac{1}{4}} \\ \\ \Rightarrow \sqrt{x^2+x+1}>x+\dfrac{1}{2},\quad \forall x\in \mathbb{R}$