Question

Usar L'Hospital em lim x^(raiz de x)
x->0

Answer 1

$L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;x^{\sqrt{x}}$


É importante observar que o limite real só existe pela direita, pois a raiz quadrada não está definida para valores negativos de $x.$


Tomando o logaritmo natural dos dois lados, temos

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


Aplicando $x\to 0^{+}$ na última linha acima, chegamos a uma indeterminação do tipo $\infty/\infty.$ Logo, podemos aplicar a Regra de L'Hopital:

$\mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{\frac{d}{dx}(\mathrm{\ell n\,}x)}{\frac{d}{dx}(\frac{1}{\sqrt{x}})}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{\frac{d}{dx}(\mathrm{\ell n\,}x)}{\frac{d}{dx}(x^{-1/2})}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{\frac{1}{x}}{-\frac{1}{2}\,x^{-3/2}}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{1}{x}\cdot \dfrac{-2}{x^{-3/2}}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{-2}{x\cdot x^{-3/2}}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{-2}{x^{1-(3/2)}}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;\dfrac{-2}{x^{-1/2}}\\ \\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;-2x^{1/2}\\ \\ \mathrm{\ell n\,}L=\underset{x\to 0^{+}}{\mathrm{\ell im}}\;-2\sqrt{x}\\ \\ \mathrm{\ell n\,}L=-2\cdot \sqrt{0}\\ \\ \mathrm{\ell n\,}L=0$


Aplicando a exponencial aos dois lados, chegamos a

$e^{\mathrm{\ell n\,}L}=e^{0}\\ \\ L=e^{0}\\ \\ L=1\\ \\ \\ \Rightarrow\;\;\boxed{\begin{array}{c} \underset{x\to 0^{+}}{\mathrm{\ell im}}\;x^{\sqrt{x}}=1 \end{array}}$