Question

$lim_{x→2} \frac{(x^{2}-2x)sen(x^{2}-4)}{\sqrt{x^{2}+4}-\sqrt{4x}}$

Answer 1

Explicação passo-a-passo:

Cálculo do Limite :

Vamos primeiramente Racionalizar e Simplificar a fracçào :

${\lim_{x \to 2} \mathsf{\dfrac{(x^2-2x).\sin(x^2-4)}{\sqrt{x^2+4}-\sqrt{4}}} } ~=~[\dfrac{0}{0}]$

$\iff \mathsf{L~=}{\lim_{x \to 2} \dfrac{(x^2-2x).\sin(x^2-4).\Big(\sqrt{x^2+4}+\sqrt{4x}\Big)}{x^2+4-4x} }$

$\iff \mathsf{L~=~}{\lim_{x \to 2} \dfrac{x\cancel{(x-2)}.\sin(x^2-4).\Big(\sqrt{x^2+4}+\sqrt{4x} \Big)}{\cancel{(x-2)}(x-2)} }$

$\iff \mathsf{L~=~}{\lim_{x \to 2} \dfrac{x\Big(\sqrt{x^2+4}+\sqrt{4x} \Big).\sin(x^2-4) }{x-2} }$

$\mathsf{L~=~} {\lim_{x \to 2} \dfrac{2.\Big(\sqrt{2^2+4}+\sqrt{4.2} \Big).\sin(x^2-4) }{x - 2} }$

$\iff \mathsf{L~=~}{\lim_{x \to 2} \dfrac{4\sqrt{8}.\sin(x^2-4)}{x-2} }$

.

Seja :

k = x² - 4

x² = k + 4

x = √(k + 4)

k → 0

$\iff \mathsf{L~=~}{\lim_{k \to 0} \dfrac{4\sqrt{8}.\sin(k)}{\sqrt{k+4}-2} }$

Vamos novamente Racionalizar o denomimador :

$\iff \mathsf{L~=~}{\lim_{k \to 0} \dfrac{4\sqrt{8}.\sin(k).\Big(\sqrt{x+4}+2 \Big)}{k+4-4} }$

$\iff \mathsf{L~=~}{\lim_{k \to 0} 4\sqrt{8}.\dfrac{\sin(k)}{k}.\Big(\sqrt{k+4}+2 \Big) }$

$\iff \mathsf{L~=~}{\lim_{k \to 0} 4\sqrt{8}.1.\Big(\sqrt{k+4}+2 \Big) }$

$\iff \mathsf{L~=~4\sqrt{8}.\Big(\sqrt{0+4}+2 \Big) }$

$\iff \mathtt{L~=~4\sqrt{8}.(2+2)~=~16\sqrt{8} }$

$\iff \boxed{\boxed{\mathtt{ \red{L~=~32\sqrt{2} } } } }$

Dúvidas??Comente!)