Question

alguém consegue ???
$\lim_{x \to \infty} (\sqrt{x+ \sqrt{x} } - \sqrt{x})$

Answer 1

$\bullet\;\; L_{0}=\underset{x \to \infty}{\mathrm{\ell im}}\;(\sqrt{x+\sqrt{x}}-\sqrt{x})$


Multiplicando e dividindo pelo conjugado $(\sqrt{x+\sqrt{x}}+\sqrt{x})$

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


$L_{0}=\dfrac{1}{\underset{x \to \infty}{\mathrm{\ell im}}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}+1}\\ \\ \\ L_{0}=\dfrac{1}{L_{1}+1}\;\;\;\;\;(i)$


onde $L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}.$


$\bullet\;\; L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}\\ \\ \\ L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{x+\sqrt{x}}{x}}\\ \\ \\ L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{x+x^{1/2}}{x}}\\ \\ \\ L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{\diagup\!\!\!\! x\cdot (1+x^{-1/2})}{\diagup\!\!\!\! x}}\\ \\ \\ L_{1}=\underset{x \to \infty}{\mathrm{\ell im}}\;\sqrt{1+x^{-1/2}}\\ \\ \\ L_{1}=\sqrt{1+\underset{x \to \infty}{\mathrm{\ell im}}x^{-1/2}}\\ \\ \\ L_{1}=\sqrt{1+0}\\ \\ \\ L_{1}=1$


Substituindo o valor de $L_{1}$ em $(i)$

$L_{0}=\dfrac{1}{1+1}\\ \\ \\ L_{0}=\dfrac{1}{2}\\ \\ \\ \Rightarrow\;\;\boxed{ \begin{array}{c} \underset{x \to \infty}{\mathrm{\ell im}}\;(\sqrt{x+\sqrt{x}}-\sqrt{x})=\dfrac{1}{2} \end{array} }$