Question

Sa se calculeze:



$\lim _{x\to \:0}\:\frac{f\left(x\right)}{g\left(x\right)}$



f(x)=x³+3x-4


g(x)=x³-5x+4

Answer 1

$\lim\limits_{x\to 1}\dfrac{x^3+3x-4}{x^3-5x+4} = \lim\limits_{x\to 1}\dfrac{(x-1)(x^2+x+4)}{(x-1)(x^2+x-4)}=\\ \\\\ = \lim\limits_{x\to 1}\dfrac{x^2+x+4}{x^2+x-4} = \dfrac{1^2+1+4}{1^1+1-4} = \dfrac{6}{-2} = \boxed{-3}$