Question

Qual é o limite ? [2x +11] \ sqrt[x+1] com x tendendo a infinito ? sem L `Hopital, pedido do professor.

Answer 1

$L=\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2x+11}{\sqrt{x+1}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2x\cdot (1+\frac{11}{2x})}{\sqrt{x\cdot (1+\frac{1}{x})}}$


Como $x\to\infty,$ temos que $x>0.$ Portanto, podemos separar as raízes no denominador assim:

$=\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2x\cdot (1+\frac{11}{2x})}{\sqrt{x}\cdot \sqrt{1+\frac{1}{x}}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2x}{\sqrt{x}}\cdot\dfrac{ 1+\frac{11}{2x}}{\sqrt{1+\frac{1}{x}}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2(\sqrt{x})^{2}}{\sqrt{x}}\cdot\dfrac{ 1+\frac{11}{2x}}{\sqrt{1+\frac{1}{x}}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~2\sqrt{x}\cdot\dfrac{ 1+\frac{11}{2x}}{\sqrt{1+\frac{1}{x}}}~~~~~~\mathbf{(i)}$


Sabemos que

$\underset{x\to \infty}{\mathrm{\ell im}}~2\sqrt{x}=\infty~~~~~~\mathbf{(ii)}\\ \\ \\ \underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{ 1+\frac{11}{2x}}{\sqrt{1+\frac{1}{x}}}=\dfrac{1+0}{\sqrt{1+0}}=1~~~~~~\mathbf{(iii)}$


Portanto, por $\mathbf{(ii)}$ e $\mathbf{(iii)},$ temos que

$\underset{x\to \infty}{\mathrm{\ell im}}~2\sqrt{x}\cdot\dfrac{ 1+\frac{11}{2x}}{\sqrt{1+\frac{1}{x}}}=\infty\\ \\ \\ \\ \Rightarrow~\boxed{\begin{array}{c} \underset{x\to \infty}{\mathrm{\ell im}}~\dfrac{2x+11}{\sqrt{x+1}}=\infty \end{array}}$