от ammornil » 22 Яну 2021, 16:13
[tex]\lim_{x \to 2}\frac{4-x^{2}}{2-x}=\lim_{x \to 2}\frac{\cancel{(2-x)}.(2+x)}{\cancel{2-x}}=...=2+2=4[/tex]
[tex]\lim_{x \to \infty}\frac{-2n^{3}+5n^{2}}{4n^{3}+1}=\lim_{x \to \infty} \frac{\cancel{n^{3}}.(-2+\frac{5}{n})}{\cancel{n^{3}}.(4+\frac{1}{n})}=...=\frac{-2+\frac{5}{\infty}}{4+\frac{1}{\infty}}=\frac{-2+0}{4+0}=\frac{-2}{4}=-\frac{1}{2}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]