По свойствам пределов:
\infty} \frac{2-3n}{n+2}=lim_{n->\infty} \frac{\frac{2}{n}-3}{1+\frac{2}{n}}=\frac{0-3}{1+0}=-3" alt="lim_{n->\infty} \frac{2-3n}{n+2}=lim_{n->\infty} \frac{\frac{2}{n}-3}{1+\frac{2}{n}}=\frac{0-3}{1+0}=-3" align="absmiddle" class="latex-formula">