Question

Sendo f(x)= 2x+5 , (fºg)(x)= 2x²-4x+7, determine lim x¬1 g(x)/7x³-f(x).

obs: o limite é g(x) sobre 7x³-f(x).

Answer 1

$f(x)=2x+5\\\\ (fog)=2x^2-4x+7$

como fog é vc substituir o x em f(x) por g(x) então
$(fog) = 2g+5 =2x^2-4x+7\\\\2g+5=2x^2-4x+7\\\\2g=2x^2-4x+7-5\\\\2g=2x^2-4x+2\\\\g= \frac{2x^2-4x+2}{2} \\\\ \boxed{g(x)=x^2-2x+1}$

calculando o limite

$\lim_{x \to 1} \frac{g(x)}{7x^3-f(x)} \\\\ \lim_{x \to 1} \frac{x^2-2x+1}{7x^3-(2x+5)} \\\\ \lim_{x \to 1} \frac{x^2-2x+1}{7x^3-2x-5}= \frac{0}{0}$

resolvendo a indeterminação 
fatorando o numerador (a-b)² = a²-2ab+b²)
então x²-2x+1 = (x-1)²

no denominador aplicando birot ruffini
$\boxed{1}\boxed{7_{x^3}}\boxed{0_{x^2}}\boxed{-2_x}\boxed{-5}\\\boxed{}\boxed{7}\boxed{7}\boxed{5}\boxed{0}\\\\\\ 7x^3-2x-5 = (x-1)*(7x^2-7x+5)$


reescrevendo o limite

$\lim_{x \to 1} \frac{x^2-2x+1}{7x^3-2x-5}\\\\\lim_{x \to 1} \frac{(x-1)^2}{(x-1)*(7x^2-7x+5)}\\\\\boxed{\boxed{ lim_{x \to 1} \frac{(x-1)}{(7x^2-7x+5)}= \frac{(1-1)}{7*1^2-7*1+5}= \frac{0}{5}=0 }}$