Question

$\lim_{x \to -3} \sqrt{ x^{2} } +16-5/ x^{2} +3x$

Answer 1

Olá

Limites, indeterminação.


$\displaystyle \mathsf{\lim_{x \to -3} \frac{ \sqrt{x^2+16} -5}{x^2+3x}~=~ \frac{0}{0} }$


Temos uma indeterminação do tipo 0/0.

Para sairmos disso temos que multiplicar pelo conjugado do numerador.


$\displaystyle \mathsf{\lim_{x \to -3} \frac{ \sqrt{x^2+16} -5}{x^2+3x} \cdot \frac{\sqrt{x^2+16} +5}{\sqrt{x^2+16} +5} }\\\\\\\mathsf{\lim_{x \to -3} \frac{ (\sqrt{x^2+16})^2 -(5)^2}{(x^2+3x)(\sqrt{x^2+16} +5)} }\\\\\\\mathsf{\lim_{x \to -3} \frac{x^2+16-25}{(x^2+3x)(\sqrt{x^2+16} +5)} }\\\\\\\mathsf{\lim_{x \to -3} \frac{x^2-9}{(x^2+3x)(\sqrt{x^2+16} +5)} }$


Agora vamos pegar o numerador e fatorar

x²-9
x²-9 = 0 
x = +-√9
x = +-3


Agora vamos fatorar o termo

x² + 3x

x² + 3x = 0
x(x + 3)


Substituindo nno limite, o que foi fatorado


$\displaystyle \mathsf{\lim_{x \to -3}~ \frac{(x-3)(x+3)}{x(x+3)(\sqrt{x^2+16} +5)} }\\\\\\\text{Simplifica}\\\\\mathsf{\lim_{x \to -3}~ \frac{(x-3)(\diagup\!\!\!\!x+\diagup\!\!\!\!3)}{x(\diagup\!\!\!\!x+\diagup\!\!\!\!3)(\sqrt{x^2+16} +5)} }\\\\\\ \mathsf{\lim_{x \to -3}\frac{x-3}{x(\sqrt{x^2+16} +5)} }\\\\\\\text{Resolve o limite}\\\\\\ \mathsf{\lim_{x \to -3}\frac{(-3)-3}{(-3)(\sqrt{(-3)^2+16} +5)}~=~\frac{-6}{(-3)( \sqrt{25}+5 )}~=~\frac{-6}{-3\cdot 10}~=~ \frac{-6}{-30}}\\\\\\\mathsf{=\frac{1}{5} }$

$\displaystyle \boxed{\mathsf{\lim_{x \to -3} \frac{ \sqrt{x^2+16} -5}{x^2+3x}~=~ \frac{1}{5} }}$




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