Question

Acest exercițiu. Vă mulțumesc!​

Answer 1

Răspuns:

1)

$\displaystyle\lim_{n\to\infty}\left(\dfrac{2^n-1}{2^n+1}\right)^{3^n}=\lim_{n\to\infty}\left(1-\dfrac{2}{2^n+1}\right)^{3^n}=\lim_{n\to\infty}\left[\left(1-\dfrac{2}{2^n+1}\right)^{-\frac{2^n+1}{2}\right]^{-\frac{2\cdot 3^n}{2^n+1}}=\\=e^{-\displaystyle\lim_{n\to\infty}\left(\frac{3}{2}\right)^n\cdot\frac{2}{1+\frac{1}{2^n}}}=e^{-\infty}=0$

2)

$\displaystyle\lim_{n\to\infty}(2n-3)(\sqrt[n]{5}-1)=\lim_{n\to\infty}n\left(2-\frac{3}{n}\right)\left(5^{\frac{1}{n}}-1\right)=\\=\lim_{n\to\infty}\dfrac{5^{\frac{1}{n}-1}}{\frac{1}{n}}\cdot\left(2-\frac{3}{n}\right)=2\ln 5$

3)

$\left|\dfrac{\sin\sqrt{n}}{\sqrt[3]{n}}\right|\le \dfrac{1}{\sqrt[3]{n}}\to 0$

Deci $\displaystyle\lim_{n\to\infty}\dfrac{\sin\sqrt{n}}{\sqrt[3]{n}}=0$

Explicație pas cu pas: