Question

Como resolver este limite ?? Alguém pode ajudar?

Answer 1

$\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{\sqrt{x}-3}{x^{2}-9x}\\ \\ =\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{\sqrt{x}-3}{x\cdot \left(x-9 \right )}\\ \\ =\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{\left(\sqrt{x}-3 \right )\cdot \left(\sqrt{x}+3 \right )}{x\cdot \left(x-9 \right )\cdot \left(\sqrt{x}+3 \right )}\\ \\ =\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{\left(\sqrt{x} \right )^{2}-3^{2}}{x\cdot \left(x-9 \right )\cdot \left(\sqrt{x}+3 \right )}\\ \\ =\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{x-9}{x\cdot \left(x-9 \right )\cdot \left(\sqrt{x}+3 \right )}$


Cancelado o fator $x-9$ no numerador e no denominador, temos

$=\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{1}{x\cdot \left(\sqrt{x}+3 \right )}\\ \\ =\dfrac{1}{9\cdot \left(\sqrt{9}+3 \right )}\\ \\ =\dfrac{1}{9\cdot \left(3+3 \right )}\\ \\ =\dfrac{1}{9\cdot 6}\\ \\ =\dfrac{1}{54}\\ \\ \\ \boxed{\underset{x \to 9}{\mathrm{\ell im}}\;\dfrac{\sqrt{x}-3}{x^{2}-9x}=\dfrac{1}{54}}$


Resposta: alternativa $\text{e) }1/54$.