Question

limite de
$\frac{ (3 - \sqrt{h^2+h+9)} }{h^3 +1}$

resultado 1/18

x tende a menos um 

Answer 1

$\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}$

Multiplicando o numerador e o denominador pelo "conjugado" do numerador, que é $3+\sqrt{h^{2}+h+9}$, ficamos com

$\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{(3-\sqrt{h^{2}+h+9})(3+\sqrt{h^{2}+h+9})}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}\\\\\\\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{3^{2}-(\sqrt{h^{2}+h+9})^{2}}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}\\\\\\\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{3^{2}-|h^{2}+h+9|}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}$

Se estudarmos a função $f(h)=h^{2}+h+9$, vemos que ela não possui raízes reais (não corta o eixo x) e seu gráfico é uma parábola com concavidade para cima. Com essas informações, temos que $f(h)=h^{2}+h+9~\textgreater~0~~\forall~h\in\mathbb{R}$

Daí, $|f(h)|=|h^{2}+h+9|=h^{2}+h+9~~\forall~h\in\mathbb{R}$

Então:

$\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{3^{2}-(h^{2}+h+9)}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}\\\\\\\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{9-h^{2}-h-9}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}\\\\\\\lim\limits_{x\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{x\to-1}\dfrac{-h^{2}-h}{(h^{3}+1)(3+\sqrt{h^{2}+h+9})}\\\\\\=\lim\limits_{x\to-1}\dfrac{-h(h+1)}{(h^{3}+1^{3})(3+\sqrt{h^{2}+h+9})}$

Usando $a^{3}+b^{3}=(a+b)\cdot(a^{2}-ab+b^{2})$ em $h^{3}+1^{3}$:

$\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{h\to-1}\dfrac{-h(h+1)}{(h+1)(h^{2}-h+1)(3+\sqrt{h^{2}+h+9})}$

Como $h\neq-1$ no limite, podemos cancelar $h+1$:

$\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\lim\limits_{h\to-1}\dfrac{-h}{(h^{2}-h+1)(3+\sqrt{h^{2}+h+9})}$

Agora não temos mais indeterminação e podemos substituir h por -1:

$\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\dfrac{-(-1)}{([-1]^{2}-[-1]+1)(3+\sqrt{[-1]^{2}+[-1]+9})}\\\\\\\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\dfrac{1}{(1+1+1)(3+\sqrt{1-1+9})}\\\\\\\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\dfrac{1}{3\cdot(3+\sqrt{9})}\\\\\\\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\dfrac{1}{3\cdot(3+3)}\\\\\\\boxed{\boxed{\lim\limits_{h\to-1}\dfrac{3-\sqrt{h^{2}+h+9}}{h^{3}+1}=\dfrac{1}{18}}}$