Question

Sobre Limite (x,y) → (0,0) x² + y² / √ x² + y² + 1 - 1

Answer 1

$\\ \lim_{(x,y) \to 0} \frac{x^2+y^2}{ \sqrt{x^2+y^2+1} -1} \\ \\ \lim_{(x,y) \to 0} \frac{x^2+y^2}{ \sqrt{x^2+y^2+1} -1} * \frac{ \sqrt{x^2+y^2+1} +1}{ \sqrt{x^2+y^2+1} +1} \\ \\ \lim_{(x,y) \to 0} \frac{(x^2+y^2)}{ \sqrt{x^2+y^2+1} -1} \frac{( \sqrt{x^2+y^2+1} +1)}{( \sqrt{x^2+y^2+1} +1)} \\ \\ \lim_{(x,y) \to 0} \frac{(x^2+y^2)(\sqrt{x^2+y^2+1} +1)}{ ( \sqrt{x^2+y^2+1})^2-1 } \\ \\ \lim_{(x,y) \to 0} \frac{(x^2+y^2)(\sqrt{x^2+y^2+1} +1)}{ x^2+y^2+1-1 }$

$\\ \lim_{(x,y) \to 0} \frac{(x^2+y^2)(\sqrt{x^2+y^2+1} +1)}{ x^2+y^2 } \\ \\ \lim_{(x,y) \to 0} \frac{1\sqrt{x^2+y^2+1} +1)}{ 1 } \\ \\ \lim_{(x,y) \to 0} \sqrt{0^2+0^2+1} +1 \\ \\ \lim_{(x,y) \to 0} \sqrt{1} +1 \\ \\ \lim_{(x,y) \to 0} 1+1 \\ \\ \lim_{(x,y) \to 0} 2 = 2$

Answer 2

Bom dia!

Solução!

$\displaystyle \lim_{(x,y) \to (0,0)} \frac{ x^{2} +y^{2} }{ \sqrt{ x^{2} +y^{2}-1 } +1}$

Vamos usar essa propriedade para resolver o limite.

$A^{2} +B^{2}=(A-B).(A+B)$


$\displaystyle \lim_{(x,y) \to (0,0)} \frac{ x^{2} +y^{2} }{ \sqrt{ x^{2} +y^{2}+1 } -1} \times \frac{ \sqrt{ x^{2} +y^{2}+1 } +1}{ \sqrt{ x^{2} +y^{2}+1 }+1 }\\\\\\ \displaystyle \lim_{(x,y) \to (0,0)} \frac{ (x^{2} +y^{2})( \sqrt{ x^{2} +y ^{2} +1} +1}{( x^{2} +y^{2}) } \\\\\\ \displaystyle \lim_{(x,y) \to (0,0)} \sqrt{ x^{2} +y^{2}+1 }+1\\\\\\ \displaystyle \lim_{(x,y) \to (0,0)} \sqrt{ 0^{2} +0^{2}+1 }+1\\\\\\ \displaystyle \lim_{(x,y) \to (0,0)} \sqrt{ 1 }+1$


$\displaystyle \lim_{(x,y) \to (0,0)} \sqrt{ 1 }+1\\\\\\ \displaystyle \lim_{(x,y) \to (0,0)} 2\\\\\\\\\\\ \boxed{\displaystyle \lim_{(x,y) \to (0,0)} \frac{ x^{2} +y^{2} }{ \sqrt{ x^{2} +y^{2}-1 } +1}=2}\\\\\\\ \boxed{Resposta:Alternativa~~c}$

Bom dia!
Bons estudos!