Question

Calculati limita sirului:
$\lim_{n \to \infty} \dfrac{2-sinn }{2n+3}$

Answer 1

$\lim _{n\to \infty }\left(\frac{2-sin\left(n\right)}{2n+3}\right)=\lim _{n\to \infty }\left(\frac{n\left(\frac{2}{n}-\frac{sin\left(n\right)}{n}\right)}{n\left(2+\frac{3}{n}\right)}\right)=\lim _{n\to \infty }\left(\frac{\frac{2}{n}-\frac{sin\left(n\right)}{n}}{2+\frac{3}{n}}\right)$$-1\le sin\left(n\right)\le 1= > -\frac{1}{n}\le \frac{sin\left(n\right)}{n}\le \frac{1}{n}= > \lim _{n\to \infty }\left(-\frac{1}{n}\right)\le \lim _{\to \infty }\left(\frac{sin\left(n\right)}{n}\right)\le \lim _{n\to \infty }\left(\frac{1}{n}\right)= > 0\le \lim \:_{\to \:\infty \:}\left(\frac{sin\left(n\right)}{n}\right)\le \:0= > \lim _n_{\to \:\:\infty \:\:}\left(\frac{sin\left(n\right)}{n}\right)=0$$\lim _{n\to \infty }\left(\frac{\frac{2}{n}-\frac{sin\left(n\right)}{n}}{2+\frac{3}{n}}\right)=\frac{0-0}{2+0}=\frac{0}{2}=0$