Question

lim      (√(x+9) - 3)/x
x⇒0               

Answer 1

$L=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{x+9}-3}{x}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{x+9}-3}{x}\cdot \dfrac{\sqrt{x+9}+3}{\sqrt{x+9}+3}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\big(\sqrt{x+9}-3\big)\cdot \big(\sqrt{x+9}+3\big)}{x\cdot \big(\sqrt{x+9}+3\big)}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\big(\sqrt{x+9}\big)^{\!2}-3^2}{x\cdot \big(\sqrt{x+9}+3\big)}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{x+9-9}{x\cdot \big(\sqrt{x+9}+3\big)}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\diagup\!\!\!\! x}{\diagup\!\!\!\! x\cdot \big(\sqrt{x+9}+3\big)}$

$=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{1}{\sqrt{x+9}+3}\\\\\\ =\dfrac{1}{\sqrt{0+9}+3}\\\\\\ =\dfrac{1}{\sqrt{9}+3}\\\\\\ =\dfrac{1}{3+3}\\\\\\ =\dfrac{1}{6}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{x+9}-3}{x}=\dfrac{1}{6} \end{array}}$


Bons estudos! :-)