Question

lim x->0 raiz quadrada(16-x)-4/x

Answer 1

$L=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{16-x}-4}{x}$


Multiplicando e dividindo pelo conjugado do numerador, o limite fica

$=\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\left(\sqrt{16-x}-4 \right )\!\left(\sqrt{16-x}+4 \right )}{x\left(\sqrt{16-x}+4 \right )}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\left(\sqrt{16-x} \right )^2-4^2}{x\left(\sqrt{16-x}+4 \right )}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{16-x-16}{x\left(\sqrt{16-x}+4 \right )}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{-\diagup\!\!\!\! x}{\diagup\!\!\!\! x\left(\sqrt{16-x}+4 \right )}\\\\\\ =\underset{x\to 0}{\mathrm{\ell im}}~\dfrac{-1}{\sqrt{16-x}+4}$

$=\dfrac{-1}{\sqrt{16-0}+4}\\\\\\ =\dfrac{-1}{\sqrt{16}+4}\\\\\\ =\dfrac{-1}{4+4}\\\\\\ =-\,\dfrac{1}{8}\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{x\to 0}{\mathrm{\ell im}}~\dfrac{\sqrt{16-x}-4}{x}=-\,\dfrac{1}{8} \end{array}}$