Question

Найдите предел функции y = f(x) при x ⇒ $x_{0}$ :
1) $f(x) = frac{2 x^{2} -4x+2}{3x-3} , x --> 1$
2) $f(x) = frac{ x^{4}-27x+ x^{3} -27 }{3x-9} , x --> 1$

Answer 1

1
lim[2(x-1)²/3(x-1)]=lim[2(x-1)/3]=2*(1-1)/3=0
2
lim[(x³(x+1)-27(x+1))/(3(x-3))]=lim[(x+1)(x-3)(x²+3x+9)/3(x-3)]=
=lim[(x+1)(x²+3x+9)/3]=2*13/3=26/3=8 2/3