Question

Como que faz esse limite ?

 

$<var>\lim_{x \to \ 9} f(x)_\frac{\sqrt{x} - 3}{x^{2}-9x}</var>$

Answer 1

$<var>\lim_{x \to \ 9} \frac{\sqrt{x} - 3}{x^{2}-9x}\\ \\ obs \ \ x^{2}-9x=x(x-9)=x(\sqrt{x} - 3)(\sqrt{x} + 3)\\ \\ \lim_{x \to \ 9} \frac{\sqrt{x} - 3}{x(\sqrt{x} - 3)(\sqrt{x} + 3)}\\ \\ \lim_{x \to \ 9} \frac{1}{x(\sqrt{x} + 3)}\\ \\ \frac{1}{9(\sqrt{9} + 3)}\\ \\ \frac{1}{9(3 + 3)}\\ \\ \frac{1}{9(6)}\\ \\ \frac{1}{54}\\</var>$

Answer 2

$\lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x^2 - 9x} = \\\\\\ \lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x^2 - 9x} \times \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \\\\\\ \lim_{x \rightarrow 9} \frac{x - 9}{x(x - 9)(\sqrt{x} + 3)} = \\\\\\ \lim_{x\rightarrow 9} \frac{1}{x(\sqrt{x} + 3)} = \\\\\\ \lim_{x \rightarrow 9} \frac{1}{9(3 + 3)} = \\\\ \boxed{\frac{1}{54}}$