Question

O valor do Limite: $\lim_{x \to \(0+} x^x$

A) 1/x
B) 0
C) lnx
D) 1
E) x

Answer 1

$L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\,x^{x}\\ \\ \\ \mathrm{\ell n\,}L=\mathrm{\ell n\,}\left(\underset{x\to 0^{+}}{\mathrm{\ell im}}\,x^{x} \right )\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\,\mathrm{\ell n\,}(x^{x})\\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\,x\,\mathrm{\ell n\,}(x)\\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\,\dfrac{\ell n\,(x)}{(\frac{1}{x})}$


O limite do lado direito é uma indeterminação do tipo $\infty/\infty.$ Aplicando a Regra de L'Hopital, temos

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