Question

Usar L' Hopital


lim (lnx)/ raiz de x
x-> infinito

Answer 1

$L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\mathrm{\ell n\,}x}{\sqrt{x}}$


Aplicando $x\to \infty,$ caímos em uma indeterminação do tipo $\infty/\infty.$ Logo, podemos aplicar a regra de L'Hopital:

$L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\frac{d}{dx}(\mathrm{\ell n\,}x)}{\frac{d}{dx}(\sqrt{x})}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\frac{d}{dx}(\mathrm{\ell n\,}x)}{\frac{d}{dx}(x^{1/2})}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{(\frac{1}{x})}{(\frac{1}{2}\,x^{-1/2})}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{1}{x}\cdot \dfrac{2}{x^{-1/2}}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{2}{x\cdot x^{-1/2}}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{2}{x^{1-(1/2)}}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{2}{x^{1/2}}\\ \\ \\ L=\underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{2}{\sqrt{x}}\\ \\ \\ L=0\\ \\ \\ \Rightarrow\;\;\boxed{\begin{array}{c} \underset{x\to \infty}{\mathrm{\ell im}}\;\dfrac{\mathrm{\ell n\,}x}{\sqrt{x}}=0 \end{array}}$