Question

Quanto é o limite ? sqrt cubica [(8 +x^2)] \ [x(x+1) ] com x tendendo a infinito

Answer 1

$L=\underset{x\to \infty}{\mathrm{\ell im}}~\,^{3}\!\!\!\!\sqrt{\dfrac{8+x^{2}}{x(x+1)}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\,^{3}\!\!\!\!\sqrt{\dfrac{x^{2}+8}{x^{2}+1}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\,^{3}\!\!\!\!\sqrt{\dfrac{\diagup\!\!\!\!\!x^{2}\cdot (1+\frac{8}{x^{2}})}{\diagup\!\!\!\!\!x^{2}\cdot (1+\frac{1}{x^{2}})}}\\ \\ \\ =\underset{x\to \infty}{\mathrm{\ell im}}~\,^{3}\!\!\!\!\sqrt{\dfrac{1+\frac{8}{x^{2}}}{1+\frac{1}{x^{2}}}}\\ \\ \\ =\,^{3}\!\!\!\!\sqrt{\dfrac{1+0}{1+0}}\\ \\ \\ =\,^{3}\!\!\!\!\sqrt{1}\\ \\ \\ =1\\ \\ \\ \\ \Rightarrow~\boxed{\begin{array}{c} \underset{x\to \infty}{\mathrm{\ell im}}~\,^{3}\!\!\!\!\sqrt{\dfrac{8+x^{2}}{x(x+1)}}=1 \end{array}}$