Question

Regra de L'hopital

Não sei como resolver esses dois problemas. Alguem poderia me ajudar? Com detalhees


1-) lim x->3
(raiz de x²-2x+6 - raiz de x²+2x-6 )/ (x²-4x+3)



2-) Lim x--->π/3 (1-2cosx)/(π-3x)

Answer 1

1)

$\displaystyle f(x)=\frac{\sqrt{x^2-2x+6}-\sqrt{x^2+2x-6}}{x^2-4x+3}\\ \\ \\ f(x)=\frac{\sqrt{x^2-2x+6}-\sqrt{x^2+2x-6}}{(x-3)(x-1)}\cdot\frac{\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}}{\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}}\\ \\ \\ f(x)=\frac{(x^2-2x+6)-(x^2+2x-6)}{(x-3)(x-1)\left(\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}\right)}\\ \\ \\ f(x)=\frac{-4(x-3)}{(x-3)(x-1)\left(\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}\right)}$

$\displaystyle f(x)\sim F(x)=\frac{-4}{(x-1)\left(\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}\right)}\\ \\ \\ \text{Entonces:} \\ \\ \lim\limits_{x\to 3}f(x)=F(3)\\ \\ \\ \boxed{\lim\limits_{x\to 3}f(x)=-\frac{1}{3}}$

Aplicando la regla de L'Hospital se tiene
$\displaystyle \lim\limits_{x\to 3}\frac{\frac{x-2}{\sqrt{x^2-2x+6}}-\frac{x+2}{\sqrt{x^2+2x-6}}}{2x-4}=\frac{-\frac{4}{6}}{2}=\boxed{-\frac{1}{3}}$


2)

$\displaystyle L=\lim\limits_{x\to\frac{\pi}{3}}\frac{1-2\cos x}{\pi-3x}\\ \\ \\ L=\lim\limits_{x\to\frac{\pi}{3}}\frac{2\sin x}{-3}\\ \\ \\ \boxed{L=-\frac{\sqrt{3}}{3}}$