Question

Como escapar da indeterminação desse cálculo de limite?





$\lim_{x\to \ 1 } \frac{x^{3}-3x^{2}+6x-4 }{x^{3}-4x^{2}+8x-5}$

Answer 1

Fatorando ! 

$\lim_{x \to 1} \frac{ x^{3} - 3x^{2} +6x-4 }{ x^{3} -4 x^{2} +8x-5}\\\\\\\\ \lim_{x \to 1} \frac{(x-1).( x^{2} -2x+4)}{(x-1).( x^{2} -3x+5)} \\\\\\ \lim_{x \to 1} \frac{ x^{2} -2x+4}{ x^{2} -3x+5}$


Agora basta substituir ... 


$\lim_{x \to 1} \frac{ 1^{2}-2.1+4 }{1^{2}-3.1+5} \\\\\\ \lim_{x \to 1} \frac{1 - 2+4}{1-3+5} \\\\\\ \lim_{x \to 1} \frac{3}{3} =1\\\\\\Ent\~ao\ temos\ que:\\\\\\ \lim_{x \to 1} \frac{ x^{3} - 3x^{2} +6x-4 }{ x^{3} -4 x^{2} +8x-5}=\boxed{\boxed{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$