Question

limite de (x+1)^1/2 / (x-1)^1/2 qundi x tende a + infinito

Answer 1

$L=\underset{x \to +\infty}{\mathrm{\ell im}}\;\dfrac{(x+1)^{1/2}}{(x-1)^{1 /2}}\\ \\ \\ L=\underset{x \to +\infty}{\mathrm{\ell im}}\left(\dfrac{x+1}{x-1} \right )^{1/2}\\ \\ \\ L=\left(\underset{x \to +\infty}{\mathrm{\ell im}}\;\dfrac{x+1}{x-1} \right )^{1/2}$


Colocando o $x$ em evidência no numerador e no denominador, temos

$L=\left(\underset{x \to +\infty}{\mathrm{\ell im}}\;\dfrac{\diagup\!\!\!\! x\cdot (1+\frac{1}{x})}{\diagup\!\!\!\!x\cdot (1-\frac{1}{x})} \right )^{1/2}\\ \\ \\ L=\left(\underset{x \to +\infty}{\mathrm{\ell im}}\;\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}} \right )^{1/2}\\ \\ \\ L=\left(\dfrac{1+0}{1-0} \right )^{1/2}\\ \\ \\ L=\left(\dfrac{1}{1} \right )^{1/2}\\ \\ \\ L=1^{1/2}\\ \\ L=1$