granica ciągu

[TwojaKotQ]

Oblicz granicę ciągu :
a)
\(an =\frac{3n-1}{2n+5}\)

b)
\(an= \frac{5n^2+3n-7}{6n^2+2n+1}\)

[irena]

a)
\(a_n=\frac{3n-1}{2n+5}=\frac{n(3-\frac{1}{n})}{n(2+\frac{5}{n})}=\frac{3-\frac{1}{n}}{2+\frac{5}{n}}\ \to\ \frac{3}{2}\)

[irena]

b)
\(a_n=\frac{5n^2+3n-7}{6n^2+2n+1}=\frac{n^2(5+\frac{3}{n}-\frac{7}{n^2})}{n^2(6+\frac{2}{n}+\frac{1}{n^2})}=\frac{5+\frac{3}{n}-\frac{7}{n^2}}{6+\frac{2}{n}+\frac{1}{n^2}}\ \to\ \frac{5}{6}\)