Question

Qual o limite de x^9 -1 / x^3 -1 quando x tende a 1 ...

Answer 1

Explicação passo-a-passo:

Cálculo do Limite

Dado o limite :

$\sf{L~=~} \lim_{x \to 1 } \sf{ \dfrac{x^9-1}{x^3-1} } ~=~\Big| \dfrac{0}{0} \Big|$

$\sf{L~=~} \lim_{x \to 1 } \sf{ \dfrac{(x^3)^3-1^3}{(x^3-1)}}$

Note que :

$\red{\boxed{\sf{ a^3 - b^3~=~(a - B)(a^2 + ab + b^2) } } }$

$\sf{L~=~} \lim_{x \to 1} \sf{ \dfrac{ \cancel{(x^3-1)}(x^6+x^3+1) }{\cancel{(x^3-1)}}}$

$\sf{L~=~} \lim_{x \to 1} \sf{ (x^6+x^3+1) } ~=~\sf{1^6 + 1^3 + 1 }$

$\sf{L~=~1+1+1 ~=~\green{3} }$

Espero ter ajudado bastante!)

Answer 2

Oie, Td Bom?!

$= lim_{x⟶1}( \frac{x {}^{9} - 1}{x {}^{3} - 1 } )$

$= lim_{x⟶1}( \frac{(x {}^{3} ) {}^{3} - 1 {}^{3} }{x {}^{3} - 1 } )$

$= lim_{x⟶1}( \frac{(x {}^{3} - 1) \: . \: (x {}^{6} + x {}^{3} + 1) }{x {}^{3} - 1} )$

$= lim_{x⟶1}(x {}^{6} + x {}^{3} + 1 )$

$= lim_{x⟶1}(x {}^{6} + x {}^{3} ) + lim_{x⟶1}(1)$

$= lim_{x⟶1}(x {}^{6} ) + lim_{x⟶1}(x {}^{3} ) + 1$

$= ( lim_{x⟶1}(x) ) {}^{6} + ( lim_{x⟶1}(x) ) {}^{3} + 1$

$= 1 {}^{6} + 1 {}^{3} + 1$

$= 1 + 1 + 1$

$= 2 + 1$

$= 3$

Att. Makaveli1996