Question

Salut, am nevoie de ajutor la ex 614(explicatii pas cu pas ).

Answer 1

$l =\lim\limits_{n\to \infty}n^{\alpha}(\sqrt[3]{n^3+bn+1}-n)= \\ \\ = \lim\limits_{n\to \infty}n^{\alpha}\Big(\sqrt[3]{n^3(1+\frac{b}{n^2}+\frac{1}{n^3}})-n}\Big) =\\ \\=\lim\limits_{n\to \infty}n^{\alpha}\cdot n\Big(\sqrt[3]{1+\frac{b}{n^2}+\frac{1}{n^3}}-1}\Big) =\\ \\= \lim\limits_{n\to \infty }n^{\alpha+1}\Big(\sqrt[3]{1+\frac{b}{n^2}+\frac{1}{n^3}}-1}\Big)=$

$= \lim\limits_{n\to \infty}\dfrac{\sqrt[3]{1+\frac{b}{n^2}+\frac{1}{n^3}}-1 }{\Big(\dfrac{1}{n}\Big)^{\alpha+1}}}=\\ \\ \dfrac{1}{n} = t \Rightarrow t\to 0 \\ \\ = \lim\limits_{t\to 0}\dfrac{\sqrt[3]{1+bt^2+t^3}-1}{t^{\alpha+1}} \overset{\frac{0}{0}}{=} \\ \\ \overset{\frac{0}{0}}{=} \lim\limits_{t\to 0}\dfrac{\frac{1}{3}(1+bt^2+t^3)^{-\frac{2}{3}}(3t^2+2bt)}{(\alpha+1)t^{\alpha}}}= \\ \\ = \lim\limits_{t\to 0}\dfrac{t\cdot \frac{1}{3}(1+bt^2+t^3)^{-\frac{2}{3}}(3t+2b)}{(\alpha+1)t^{\alpha}}}$

$\Rightarrow \alpha = 1 \text{ ca sa se simplifice t cu t}\\ \\ \Rightarrow \lim\limits_{t\to 0}\dfrac{ \frac{1}{3}(1+bt^2+t^3)^{-\frac{2}{3}}(3t+2b)}{(1+1)} = \dfrac{\frac{1}{3}(1+0+0)^{-\frac{2}{3}}(0+2b)}{2}= \\ \\ = \dfrac{\dfrac{2b}{3}}{2} =\dfrac{b}{3} \\ \\ \Rightarrow b = 3 \\ \\ S = 1+3 \Rightarrow \boxed{S = 4}$