Question

Qual é o limite quando x->3 de [(x+6)^1/2 -3]/[x-3]?

Answer 1

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Calcular o limite:

$\mathsf{\underset{x\to 3}{\ell im}~\dfrac{\sqrt{x+6}-3}{x-3}}\\\\\\ =\mathsf{\underset{x\to 3}{\ell im}~\dfrac{\sqrt{x+6}-3}{x-3}\cdot \dfrac{\sqrt{x+6}+3}{\sqrt{x+6}+3}}\\\\\\ =\mathsf{\underset{x\to 3}{\ell im}~\dfrac{\big(\sqrt{x+6}-3\big)\cdot \big(\sqrt{x+6}+3\big)}{(x-3)\cdot \big(\sqrt{x+6}+3\big)}}$

$=\mathsf{\underset{x\to 3}{\ell im}~\dfrac{\big(\sqrt{x+6}\big)^2-3^2}{(x-3)\cdot \big(\sqrt{x+6}+3\big)}}\\\\\\ =\mathsf{\underset{x\to 3}{\ell im}~\dfrac{x+6-9}{(x-3)\cdot \big(\sqrt{x+6}+3\big)}}\\\\\\ =\mathsf{\underset{x\to 3}{\ell im}~\dfrac{x-3}{(x-3)\cdot \big(\sqrt{x+6}+3\big)}}\\\\\\ =\mathsf{\underset{x\to 3}{\ell im}~\dfrac{1}{\sqrt{x+6}+3}}$

$=\mathsf{\dfrac{1}{\sqrt{3+6}+3}}\\\\\\ =\mathsf{\dfrac{1}{\sqrt{9}+3}}\\\\\\ =\mathsf{\dfrac{1}{3+3}}\\\\\\ =\mathsf{\dfrac{1}{6}}$


$\therefore~~\boxed{\begin{array}{c} \mathsf{\underset{x\to 3}{\ell im}~\dfrac{\sqrt{x+6}-3}{x-3}=\dfrac{1}{6}} \end{array}}\qquad\quad\checkmark$


Bons estudos! :-)


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