Question

lim (x3-x2-5x-3)/(x+1)2 quando x > -1

Answer 1

$\lim _{x\to \:-1}\left(\frac{x^3-x^2-5x-3}{\left(x+1\right)^2}\right)$
Fatore com briot ruffini e dps fatoração normal
$\lim _{x\to \:-1}\left(\frac{\left(x-3\right)\left(x+1\right)^2}{\left(x+1\right)^2}\right)$
$\lim _{x\to \:-1}\left(x-3\right)$
$\lim _{x\to \:-1}\left(x-3\right)$
substituie dá -1-3= -4

Answer 2

$\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}$

(Resolvi de maneira análoga, então editei e fiz de outro modo, deixando a imagem da divisão pelo Algoritmo de Briot-Ruffini)

Note que, se fizermos uma substituição, encontramos a indeterminação 0/0. Portanto, podemos usar a Regra de L'Hospital e resolver o limite:

$\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}\dfrac{\frac{d}{dx}(x^{3}-x^{2}-5x-3)}{\frac{d}{dx}(x+1)^{2}}\\\\\\\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}\dfrac{3x^{2}-2x-5}{2\cdot(x+1)^{1}\frac{d}{dx}(x+1)}\\\\\\\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}\dfrac{3x^{2}-2x-5}{2x+2}$

Novamente, se tentarmos fazer substituição, chegamos na indeterminação 0/0

Usando a Regra de L'Hospital novamente:

$\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}\dfrac{\frac{d}{dx}(3x^{2}-2x-5)}{\frac{d}{dx}(2x+2)}\\\\\\\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}\dfrac{6x-2}{2}\\\\\\\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=\lim\limits_{x\rightarrow-1}(3x-1)\\\\\\\lim\limits_{x\rightarrow-1}\dfrac{x^{3}-x^{2}-5x-3}{(x+1)^{2}}=3(-1)-1$


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