Question

lim x_1 2x²+4x-6/ x²+3x-4

Answer 1

$\displaystyle \lim_{x\to1}\frac{2x^2+4x+6}{x^2+3x-4}$
i) tentar calcular limite sem fatorar:
$\displaystyle \lim_{x\to1}\frac{2x^2+4x+6}{x^2+3x-4}=\frac{2\cdot1^2+4\cdot1-6}{1^2+3\cdot1-4}=\frac{2+4-6}{1+3-4}=\frac{0}{0}=\infty$

I) POR FATORAÇÂO:

ii) fatorar as funções de cima e a de baixo:
$\displaystyle \lim_{x\to1}\frac{2x^2+4x-6}{x^2+3x-4}\\\\ 2x^2+4x-6=0\implies X=\frac{-4\pm\sqrt{4^2-4\cdot2\cdot-6}}{2\cdot2}=\frac{-4\pm\sqrt{64}}{4}\\\\ \left \{ {{x'=\frac{-4+8}{4}=1} \atop {x''=\frac{-4-8}{4}=-3}} \right. \implies \boxed{2(x-1)(x+3)=2x^2+4x-6}$
$\displaystyle \lim_{x\to1}\frac{2x^2+4x-6}{x^2+3x-4}\\\\ x^2+3x-4=0\\\\ X=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot-4}}{2}=\frac{-3\pm\sqrt{25}}{2}= \left \{ {{x'= \frac{-3+5}{2}=1 } \atop {x''= \frac{-3-5}{2}=-4 }} \right. \\\\ \implies (x-1)(x+4)=x^2-3x-4$
$\displaystyle \lim_{x\to1}\frac{2x^2+4x-6}{x^2+3x-4}=\lim_{x\to1}\frac{2(x-1)(x+3)}{(x-1)(x+4)}=\lim_{x\to1}2\frac{x+3}{x+4}=2\frac{1+3}{1+4}=\boxed{\frac{8}{5}}$

ii) L'Hôpital
$\displaystyle \lim_{x\to1}\frac{2x^2+4x-6}{x^2+3x-4}=\lim_{x\to1}\frac{(2x^2+4x-6)'}{(x^2+3x-4)'}=\lim_{x\to1}\frac{4x+4}{2x+3}=\frac{4\cdot1+4}{2\cdot1+3}=\boxed{\frac{8}{5}}$

lembrando L'Hôpital só serve para indeterminações do tipo 0/0 ou ∞/∞.