Question

Qual o limite de (lnx)^(1/x) com x tendendo ao infinito?
Uasando a regra de l'hospital

Answer 1

$L=\underset{x \to \infty}{\mathrm{\ell im}}\left(\left(\mathrm{\ell n\,}x \right)^{1/x} \right )\\ \\ \\ \mathrm{\ell n\,}L=\mathrm{\ell n\,}\left[\underset{x \to \infty}{\mathrm{\ell im}}\left(\left(\mathrm{\ell n\,}x \right)^{1/x} \right ) \right ]\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\left[\mathrm{\ell n}\left(\left(\mathrm{\ell n\,}x \right)^{1/x} \right ) \right ]\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\left[\dfrac{1}{x}\cdot \mathrm{\ell n}\left(\mathrm{\ell n\,}x \right) \right ]\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\,\dfrac{\mathrm{\ell n}\left(\mathrm{\ell n\,}x \right)}{x}$


O limite do lado direito da igualdade é do tipo $\infty / \infty$. Logo, pela Regra de L'Hospital podemos calcular o limite do quociente entre a derivadas do numerador e a derivada do denominador, em relação a $x$:

$\mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\,\dfrac{\frac{d}{dx}\left[\mathrm{\ell n}\left(\mathrm{\ell n\,}x \right) \right]}{\frac{d}{dx}\left(x \right )}\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\,\dfrac{\frac{1}{\mathrm{\ell n\,}x}\cdot \frac{d}{dx}\left(\mathrm{\ell n\,}x \right )}{1}\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\left[\dfrac{1}{\mathrm{\ell n\,}x}\cdot \dfrac{1}{x} \right ]\\ \\ \mathrm{\ell n\,}L=\underset{x \to \infty}{\mathrm{\ell im}}\,\dfrac{1}{x\,\mathrm{\ell n\,}x}$


O denominador tende a infinito, enquanto o numerador é uma constante. Logo, o limite é igual a zero.

$\mathrm{\ell n\,}L=0\\ \\ L=e^{0}\\ \\ L=1\;\;\Rightarrow\;\;\boxed{ \underset{x \to \infty}{\mathrm{\ell im}}\left(\left(\mathrm{\ell n\,}x \right)^{1/x} \right )=1 }$