Question

sobre limite
raiz de x+3 -raiz 3 dividido por x lim 0

Answer 1

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}$

Vamos multiplicar o numerador e o denominador pelo conjugado do denominador:

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\lim\limits_{x\rightarrow0}\dfrac{(\sqrt{x+3}-\sqrt{3})(\sqrt{x+3}+\sqrt{3})}{x(\sqrt{x+3}+\sqrt{3})}\\\\\\\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\lim\limits_{x\rightarrow0}\dfrac{(\sqrt{x+3})^{2}-(\sqrt{3})^{2}}{x(\sqrt{x+3}+\sqrt{3})}\\\\\\\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\lim\limits_{x\rightarrow0}\dfrac{x+3-3}{x(\sqrt{x+3}+\sqrt{3})}$

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\lim\limits_{x\rightarrow0}\dfrac{x}{x(\sqrt{x+3}+\sqrt{3})}$

Cortando x:

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\lim\limits_{x\rightarrow0}\dfrac{1}{\sqrt{x+3}+\sqrt{3}}$

A função é contínua em x = 0, portanto, podemos fazer a substituição direta:

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\dfrac{1}{\sqrt{0+3}+\sqrt{3}}\\\\\\\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\dfrac{1}{\sqrt{3}+\sqrt{3}}\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\dfrac{1}{2\sqrt{3}}}}$

Racionalizando:

$\boxed{\boxed{\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+3}-\sqrt{3}}{x}=\dfrac{\sqrt{3}}{6}}}$