Question

questoes de limites. como resolver​

Answer 1

$\lim_{x \to 2} \dfrac{\sqrt{2x^2+3x+2}-4}{8-4x}$

Racionalize o numerador:

$\lim_{x \to 2} \dfrac{\sqrt{2x^2+3x+2}-4}{8-4x} \cdot \dfrac{\sqrt{2x^2+3x+2}+4}{\sqrt{2x^2+3x+2}+4}\\\\\lim_{x \to 2} \dfrac{2x^2+3x+2-4^2}{(8-4x)(\sqrt{2x^2+3x+2}+4)}\\\\\lim_{x \to 2} \dfrac{2x^2+3x-14}{(8-4x)(\sqrt{2x^2+3x+2}+4)}$

Fatore o numerador e (8-4x):

$2x^2+3x-14=0\\\\x=\dfrac{-3\pm \sqrt{3^2-4 \cdot 2 \cdot (-14)}}{2 \cdot 2} = \dfrac{-3\pm\sqrt{9-4 \cdot (-28)}}{4} = \dfrac{-3\pm\sqrt{121}}{4}\\\\x_1 = \dfrac{-3+11}{4} = 2\\\\x_2 = \dfrac{-3-11}{4} = \dfrac{-14}{4} = -\dfrac{7}{2}$

$\lim_{x \to 2} \dfrac{2\left(x+\dfrac{7}{2}\right)(x-2)}{4(2-x)(\sqrt{2x^2+3x+2}+4)}\\\\\\lim_{x \to 2} \dfrac{-2\left(x+\dfrac{7}{2}\right)}{4(\sqrt{2x^2+3x+2}+4)}\\\\\lim_{x \to 2} \dfrac{-2x-7}{4(\sqrt{2x^2+3x+2}+4)}\\\\\lim_{x \to 2} \dfrac{-2x-7}{4\sqrt{2x^2+3x+2}+16}$

Por substituição direta:

$\lim_{x \to 2} \dfrac{-(2\cdot 2)-7}{4\sqrt{2\cdot 2^2+3\cdot 2+2}+16}\\\\\lim_{x \to 2} \dfrac{-4-7}{4\sqrt{2\cdot 4+ 6+2}+16}\\\\\lim_{x \to 2} \dfrac{-11}{4\sqrt{8+ 8}+16}\\\\\lim_{x \to 2} \dfrac{-11}{4\sqrt{16}+16}\\\\\lim_{x \to 2} \dfrac{-11}{4\cdot 4+16}\\\\\lim_{x \to 2} \dfrac{-11}{16+16}=-\dfrac{11}{32}\\\\\\\boxed{\boxed{\lim_{x \to 2}\dfrac{\sqrt{2x^2+3x+2}-4}{8-4x} = -\dfrac{11}{32}}}$