Question

O valor do Limite lim x^x quando x tende a 0+ é:
a. 0
b. 1
c. x
d. 1/x
e. lnx

Answer 1

$L=\underset{x\to 0^+}{\mathrm{\ell im}}~x^x\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~(e^{\mathrm{\ell n\,}x})^x\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~e^{x\,\mathrm{\ell n\,}x}\\\\\\ =e^{~\underset{x\to 0^+}{\mathrm{\ell im}}x\,\mathrm{\ell n\,}x}\\\\ =e^{L_1}$

onde $L_1=\underset{x\to 0^+}{\mathrm{\ell im}}~x\,\mathrm{\ell n\,}x.$

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Calculando $L_1:$

$L_1=\underset{x\to 0^+}{\mathrm{\ell im}}~x\,\mathrm{\ell n\,}x\\\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~\dfrac{\mathrm{\ell n\,}x}{\frac{1}{x}}~~~~~~\mathbf{(i)}$


Fazendo $x\to0^+\,,$ obtemos uma indeterminação do tipo $\infty/\infty.$ Aplicando a Regra de L'Hopital o limite $\mathbf{(i)}\,,$ acima fica

(se o limite abaixo existir)

$=\underset{x\to 0^+}{\mathrm{\ell im}}~\dfrac{\frac{d}{dx}(\mathrm{\ell n\,}x)}{\frac{d}{dx}\left(\frac{1}{x} \right )}\\\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~\dfrac{\frac{1}{x}}{-\frac{1}{x^2}}\\\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~\dfrac{1}{x}\cdot (-x^2)\\\\\\ =\underset{x\to 0^+}{\mathrm{\ell im}}~(-x)\\\\ =0\\\\\\ \therefore~~L_1=0$

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Sendo assim,

$L=e^{L_1}\\\\ =e^{0}\\\\ =1\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{x\to 0^+}{\mathrm{\ell im}}~x^x=1 \end{array}}$


Resposta: alternativa $\text{b. }1.$