Question

Limx-->-1 x^3+3x^2-x-3/x^3-x^2+2​

Answer 1

$\large\boxed{\begin{array}{l}\rm x^3+3x^2-x-3= x^3+2x^2+x^2+2x-3x-3\\\rm x^3+3x^2-x-3= x^3+x^2+2x^2+2x-3x-3\\\rm x^3+3x^2-x-3= x^2(x+1)+2x(x+1)-3(x+1)\\\rm x^3+3x^2-x-3=(x+1)\cdot(x^2+2x-3)\,\boxed{1}\end{array}}$

$\large\boxed{\begin{array}{l}\rm x^3-x^2+2=x^3+x^2-2x^2+2\\\rm x^3-x^2+2=x^2(x+1)-2(x^2-1)\\\rm x^3-x^2+2=x^2(x+1)-2(x+1)(x-1)\\\rm x^3-x^2+2=(x+1)(x^2-2x+2)\,\boxed{2}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\lim_{x \to -1}\dfrac{ x^3+3x^2-x-3}{x^3-x^2+2}\\\\\sf substituindo\,\boxed{1}\,e\,\boxed{2}\,na\,express\tilde ao\,temos:\\\displaystyle\rm\lim_{x \to -1}\dfrac{\diagup\!\!\!\!(x+\diagup\!\!\!1)\cdot(x^2+2x-3)}{\diagup\!\!\!\!(x+\diagup\!\!\!\!1)(x^2-2x+2)}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\lim_{x \to -1}\dfrac{x^2+2x-3}{x^2-2x+2}=\dfrac{(-1)^2+2\cdot(-1)-3}{(-1)^2-2\cdot(-1)+2}\\\\\rm =\dfrac{1-2-3}{1+2+2}=-\dfrac{4}{5}\end{array}}$

$\large\boxed{\boxed{\boxed{\boxed{\displaystyle\rm\lim_{x \to -1}\dfrac{ x^3+3x^2-x-3}{x^3-x^2+2}=-\dfrac{4}{5}}}}}$