Question

lim (x^9-1)/(x^3-1 )
x->1

Answer 1

Explicação passo-a-passo:

$\sf lim_{x\to~1}~\dfrac{x^9-1}{x^3-1}$

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

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

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

$\sf =1^6+1^3+1$

$\sf =1+1+1$

$\sf =3$

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