Question

Aratati ca ∀n≥4,
$2^{n} \geq \frac{n^3}{24}$ si deduceti $\lim_{n \to \infty} \frac{n^2}{2^n}$ . Mersi

Answer 1

Salut,

$\dfrac{n^3}{24}\leqslant 2^n\bigg|\cdot\dfrac{24}{2^n}\Rightarrow\dfrac{n^3}{2^n}\leqslant24\bigg|:n\Rightarrow\dfrac{n^2}{2^n}\leqslant\dfrac{24}n;\lim\limits_{n\to +\infty}\dfrac{24}n=0;\;\dfrac{n^2}{2^n}>0\Rightarrow\\\\\Rightarrow Conform\;criteriului\;CLE\c{S}TELUI:\;\lim\limits_{n\to +\infty}\dfrac{n^2}{2^n}=0.$

Green eyes.

Answer 2

Folosind teorama de cleste, ne rezulta limita = 0.