• Matéria: Matemática
  • Autor: israelbdf
  • Perguntado 9 anos atrás

Alguém consegue me ajudar nessa questão .

Anexos:

carlosmath: http://www.wolframalpha.com/input/?i=lim%28%28%28n%2B1%29%2F%28n%2B3%29%29%5E%28n%2B1%29%2Cn%2C%2Binf%29

Respostas

respondido por: carlosmath
1
\displaystyle
L=\lim\limits_{n\to +\infty}\left(\frac{n+1}{n+3}\right)^{n+1}\\ \\
L=\lim\limits_{n\to +\infty}\left(1-\frac{2}{n+3}\right)^{n+1}\\ \\
L=\lim\limits_{n\to +\infty}\left(1+\dfrac{1}{-\dfrac{n+3}{2}}\right)^{-\dfrac{n+3}{2}\cdot \dfrac{2(n+1)}{n+3}}\\ \\
L=\lim\limits_{m\to -\infty}\left(1+\dfrac{1}{m}\right)^{m\cdot \lim\limits_{n\to+\infty}\dfrac{2(n+1)}{n+3}}

L=(1/e)^{\lim\limits_{n\to+\infty}\dfrac{2(n+1)}{n+3}}\\ \\ L=(1/e)^{\lim\limits_{n\to+\infty}\dfrac{2(1+\dfrac{1}{n})}{1+\dfrac{3}{n}}}\\ \\ L=(1/e)^{\dfrac{2(1+0)}{1+0}}\\ \\ \boxed{L=e^{-2}}
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