Rozwiązane

oblicz granicę ciągu:

[tex]\frac{2^{n+1}-3^{n+2} }{3^{n+2} }[/tex]

Odpowiedź :

Konrad509viz

[tex]\displaystyle\\\lim_{n\to\infty}\dfrac{2^{n+1}-3^{n+2}}{3^{n+2}}=\\\\\lim_{n\to\infty}\left(\dfrac{2^{n+1}}{3^{n+2}}-1\right)=\\\\\lim_{n\to\infty}\dfrac{2^{n+1}}{3^{n+2}}-1=\\\\\lim_{n\to\infty}\dfrac{2^n\cdot 2}{3^n\cdot3^2}-1=\\\\\dfrac{2}{9}\lim_{n\to\infty}\left(\dfrac{2}{3}\right)^n-1=\\\\\dfrac{2}{9}\cdot0-1=-1[/tex]

Viz Inne Pytanie