Question

lim x---> -3

x² + 5x + 6
________

x² - x - 12

Answer 1

$\mathrm{\lim\limits_{x\to-3}\bigg[\dfrac{x^2+5x+6}{x^2-x-12}\bigg]}$

$\mathbf{*\ Fatorando\ x^2+5x+6:}\\\\ \mathrm{x^2+5x+6=0\ \to\ x=\dfrac{-5\pm\sqrt{5^2-4.1.6}}{2.1}=\dfrac{-5\pm\sqrt{25-24}}{2}=}\\\\ \mathrm{=\dfrac{-5\pm\sqrt{1}}{2}=\dfrac{-5\pm1}{2}\ \to\ x=-3\ \ ou\ \ x=-2}\\\\ \mathrm{x^2+5x+6=(x-(-3))(x-(-2))=\boxed{\mathrm{(x+3)(x+2)}}}$

$\mathbf{*\ Fatorando\ x^2-x-12:}\\\\ \mathrm{x^2-x-12=0\ \to\ x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4.1.(-12)}}{2.1}=}\\\\ \mathrm{=\dfrac{1\pm\sqrt{1+48}}{2}=\dfrac{1\pm\sqrt{49}}{2}=\dfrac{1\pm7}{2}\ \to\ x=4\ \ ou\ \ x=-3}\\\\ \mathrm{x^2-x-12=(x-4)(x-(-3))=\boxed{\mathrm{(x-4)(x+3)}}}$

$\mathbf{*\ Calculando\ o\ limite:}\\\\ \mathrm{\lim\limits_{x\to-3}\bigg[\dfrac{x^2+5x+6}{x^2-x-12}\bigg]=\lim\limits_{x\to-3}\bigg[\dfrac{(x+3)(x+2)}{(x-4)(x+3)}\bigg]=}\\\\ \mathrm{=\lim\limits_{x\to-3}\bigg[\dfrac{x+2}{x-4}\bigg]=\dfrac{-3+2}{-3-4}=\dfrac{-1}{-7}=\boxed{\mathbf{\dfrac{1}{7}}}}$