Question

POR FAVOR! AJUDAAA!!!
Por favor, alguém me ajuda com essa questão
Calcule o seguinte limite:
$\lim_{x \to \ 1} [ \frac{1}{1-x} - \frac{3}{1 - x^{3} } ]$

Answer 1

Olá!

$\\ \displaystyle \mathsf{\lim_{x \to 1} \left [ \frac{1}{1 - x} - \frac{3}{1 - x^3} \right ] =} \\\\\\ \mathsf{\lim_{x \to 1} \left [ \frac{1}{(1 - x)} - \frac{3}{(1 - x)(1 + x + x^2)} \right ] =} \\\\\\ \mathsf{\lim_{x \to 1} \frac{1 \cdot (1 + x + x^2) - 3 \cdot 1}{(1 - x)(1 + x + x^2)} =}$

$\\ \displaystyle \mathsf{\lim_{x \to 1} \frac{1 + x + x^2 - 3}{(1 - x)(1 + x + x^2)} =} \\\\\\ \mathsf{\lim_{x \to 1} \frac{x^2 + x - 2}{(1 - x)(1 + x + x^2)} =} \\\\\\ \mathsf{\lim_{x \to 1} \frac{(x - 1)(x + 2)}{(1 - x)(1 + x + x^2)} =}$
 
 Multiplicando o numerador e o denominador por (- 1),

$\\ \displaystyle \mathsf{\lim_{x \to 1} \frac{(x - 1)(x + 2)}{(1 - x)(1 + x + x^2)} \qquad \times \frac{(- 1}{(- 1}=} \\\\\\ \mathsf{\lim_{x \to 1} \frac{- (x - 1)(x + 2)}{- (1 - x)(1 + x + x^2)}=} \\\\\\ \mathsf{\lim_{x \to 1} \frac{- (x - 1)(x + 2)}{(x - 1)(1 + x + x^2)}=}$
 
 Simplificando,

$\\ \displaystyle \mathsf{\lim_{x \to 1} - \frac{(x - 1)(x + 2)}{(x - 1)(1 + x + x^2)}=} \\\\\\ \mathsf{\lim_{x \to 1} - \frac{(x + 2)}{(1 + x + x^2)}=} \\\\\\ \mathsf{- \frac{1 + 2}{1 + 1 + 1} =} \\\\\\ \mathsf{- \frac{3}{3} =} \\\\\\ \boxed{\mathsf{- 1}}$


$\mathsf{Obs.: \ a^3 - b^3 = (a - b)(a^2 + ab + b^2)}$