justifique o porquê da sequencia an=12n+3/6n+1 convergir para 2
sidneylinkshado:
compreendi
Soluções para a tarefa
Respondido por
1
Uma sequência converge de existe um limite L real com n tendendo a infinito.
![a_n=\dfrac{12n+3}{6n+1} a_n=\dfrac{12n+3}{6n+1}](https://tex.z-dn.net/?f=a_n%3D%5Cdfrac%7B12n%2B3%7D%7B6n%2B1%7D)
Basta achar o limite:
![\lim\limits_{n\to\infty}\dfrac{12n+3}{6n+1}=\lim\limits_{n\to\infty}\dfrac{n(12+\frac{3}{n})}{n(6+\frac{1}{n})}=\lim\limits_{n\to\infty}\dfrac{12+\frac{3}{n}}{6+\frac{1}{n}}=\dfrac{12+0}{6+0}=\dfrac{12}{6}=2 \lim\limits_{n\to\infty}\dfrac{12n+3}{6n+1}=\lim\limits_{n\to\infty}\dfrac{n(12+\frac{3}{n})}{n(6+\frac{1}{n})}=\lim\limits_{n\to\infty}\dfrac{12+\frac{3}{n}}{6+\frac{1}{n}}=\dfrac{12+0}{6+0}=\dfrac{12}{6}=2](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7B12n%2B3%7D%7B6n%2B1%7D%3D%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7Bn%2812%2B%5Cfrac%7B3%7D%7Bn%7D%29%7D%7Bn%286%2B%5Cfrac%7B1%7D%7Bn%7D%29%7D%3D%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7B12%2B%5Cfrac%7B3%7D%7Bn%7D%7D%7B6%2B%5Cfrac%7B1%7D%7Bn%7D%7D%3D%5Cdfrac%7B12%2B0%7D%7B6%2B0%7D%3D%5Cdfrac%7B12%7D%7B6%7D%3D2)
Portanto, a sequencia converge para 2.
Basta achar o limite:
Portanto, a sequencia converge para 2.
Perguntas interessantes