Question

Calcule $\lim_{n \to \infty} \frac{ \frac{n+1}{ (n+1)^{2} +1} }{ \frac{n}{ n^{2}+1 } }$

Answer 1

$\large\begin{array}{l} \textsf{Calculando o limite:}\\\\ \mathsf{\underset{n\to \infty}{\ell im}~\dfrac{~~\frac{n+1}{(n+1)^2+1}~~}{\frac{n}{n^2+1}}}\\\\ =\mathsf{\underset{n\to \infty}{\ell im}~\dfrac{n+1}{(n+1)^2+1}\cdot \dfrac{n^2+1}{n}}\\\\ =\mathsf{\underset{n\to \infty}{\ell im}~\dfrac{n+1}{n^2+2n+1+1}\cdot \dfrac{n^2+1}{n}} \end{array}$

$\large\begin{array}{l} =\mathsf{\underset{n\to \infty}{\ell im}~\dfrac{n+1}{n^2+2n+2}\cdot \dfrac{n^2+1}{n}}\\\\ =\mathsf{\underset{n\to \infty}{\ell im}~\dfrac{\diagup\!\!\!\! n\cdot \left(1+\frac{1}{n}\right)}{\diagdown\!\!\!\!\! n^2\cdot \left(1+\frac{2}{n}+\frac{2}{n^2}\right)}\cdot \dfrac{\diagdown\!\!\!\!\! n^2\cdot \left(1+\frac{1}{n^2}\right)}{\diagup\!\!\!\! n}} \end{array}$

$\large\begin{array}{l} =\mathsf{\underset{n\to \infty}{\ell im}~\dfrac{1+\frac{1}{n}}{1+\frac{2}{n}+\frac{2}{n^2}}\cdot \left(1+\dfrac{1}{n^2}\right)}\\\\\vdots\\\\ =\mathsf{\dfrac{1+0}{1+0+0}\cdot (1+0)}\\\\ =\mathsf{\dfrac{1}{1}\cdot 1}\\\\ =\mathsf{1\qquad\checkmark} \end{array}$


$\large\begin{array}{l} \therefore~~\boxed{\begin{array}{l} \mathsf{\underset{n\to \infty}{\ell im}~\dfrac{~~\frac{n+1}{(n+1)^2+1}~~}{\frac{n}{n^2+1}}=1} \end{array}} \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$