Question

lim((x^3+1)/(x^2-1) as x->-1)

Answer 1

$\lim\limits_{x \to -1}\frac{x^3+1}{x^2-1}=\frac{(-1)^3+1}{(-1)^2-1}=\frac{-1+1}{1-1}=\frac00=\\ \|x^3+1=(x+1)(x^2-x+1)=x^3-x^2+x+x^2-x+1=x^3+1\|\\ \|x^2-1=(x+1)(x-1) \|\\ =\lim\limits_{x \to -1}\frac{(x+1)(x^2-x+1)}{(x-1)(x+1)}=\lim\limits_{x \to -1}\frac{(x^2-x+1)}{(x-1)}=\frac{(-1)^2-(-1)+1}{(-1)-1}=\\ =\frac{1+1+1}{-1-1}=\frac3{-2}=-\frac32=-1\frac12.$
или о правилу лопиталя
$\lim\limits_{x \to -1}\frac{x^3+1}{x^2-1}=\frac{0}{0}\stackrel{\Lambda}{=}\lim\limits_{x \to -1}\frac{\left(x^3+1\right)'}{\left(x^2-1\right)'}=\frac{3x^2}{2x}=\frac{3\cdot(-1)^2}{2\cdot(-1)}=\\ \\ \\ =\frac{3\cdot1}{-2}=\frac{3}{-2}=-\frac32=-1\frac12.$