Question

como resolve isso

$\lim_{x \to \infty} \frac{ \sqrt{x} }{ \sqrt{x + \sqrt{x + \sqrt{x} } } }$

Answer 1

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


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


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


onde $L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\sqrt{x+\sqrt{x}}}{x}.$


$\bullet\;\; L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\sqrt{x+\sqrt{x}}}{x}$

Como $x\to \infty$ ($x$ tende ao infinito positivo), então podemos escrever

$x=|x|=\sqrt{x^{2}}$

Portanto, temos que

$L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x^{2}}}\\ \\ \\ L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{x+\sqrt{x}}{x^{2}}}\\ \\ \\ L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{x+x^{1/2}}{x^{2}}}\\ \\ \\ L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\sqrt{\dfrac{\diagup\!\!\!\! x^{2}\,(x^{-1} +x^{-3/2})}{\diagup\!\!\!\! x^{2}}}\\ \\ \\ L_{2}=\underset{x\to \infty}{\mathrm{\ell im}}\;\sqrt{(x^{-1}+x^{-3/2})}\\ \\ \\ L_{2}=\sqrt{\underset{x\to \infty}{\mathrm{\ell im}}\;(x^{-1}+x^{-3/2})}\\ \\ \\ L_{2}=\sqrt{L_{3}} \;\;\;\;\;(iii)$


onde $L_{3}=\underset{x\to \infty}{\mathrm{\ell im}}\;(x^{-1}+x^{-3/2}).$


$\bullet\;\; L_{3}=\underset{x\to \infty}{\mathrm{\ell im}}\;(x^{-1}+x^{-3/2})\\ \\ L_{3}=\underset{x\to \infty}{\mathrm{\ell im}}\;x^{-1}+\underset{x\to \infty}{\mathrm{\ell im}}\;x^{-3/2}\\ \\ \\ L_{3}=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{1}{x}+\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{1}{x^{3/2}}\\ \\ \\ L_{3}=0+0\\ \\ L_{3}=0$


Substituindo o valor de $L_{3}$ em $(iii)$

$L_{2}=\sqrt{0}\\ \\ L_{2}=0$


Substituindo o valor de $L_{2}$ em $(ii)$

$L_{1}=\dfrac{1}{1+0}\\ \\ \\ L_{1}=\dfrac{1}{1}\\ \\ \\ L_{1}=1$


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

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