Question

Calcule o limite:
$\lim_{x \to \(-3} \frac{ \sqrt{x^2+16}-5 }{x^2+3x}$

A resposta é $\frac{1}{5}$ mas não consigo chegar nela

Answer 1

vc pode reescrever a raiz assim
$\sqrt{ x^{2} +16} = (x^{2} +16) ^{ \frac{1}{2} }$
depois deriva a de cima e a de baixo
$G(x) = (x^{2} +16) ^{ \frac{1}{2} } -5 \\\\G'(x) = \frac{1}{2} (x^{2} +16) ^{ \frac{1-1}{2} } .2x\\\\G'(x) = \frac{2x}{2} (x^{2} +16) ^{ \frac{-1}{2} } \\\\G'(x) = x (x^{2} +16) ^{ \frac{-1}{2} } \\\\ G'(x)=\frac{x}{ \sqrt{ x^{2}+16 } }$

derivando a de baixo
$H(x) = x^{2} +3x\\\\H'(x)=2x+3$

agora temos 

$\frac{ \frac{x}{ \sqrt{ x^{2} +16} } }{2x+3}\\\\ \frac{ \frac{-3}{ \sqrt{ -3^{2} +16} } }{(2(-3))+3}\\\\ \frac{ \frac{-3}{ \sqrt{ 25} } }{-3}\\\\ \frac{1}{ \sqrt{25} } = \frac{1}{5}$

Answer 2

 Outra,

$\lim_{x\rightarrow-3}\frac{\sqrt{x^2+16}-5}{x^2+3x}=\\\\\\\lim_{x\rightarrow-3}\frac{\sqrt{x^2+16}-5}{x^2+3x}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5}=$

$\lim_{x\rightarrow-3}\frac{(x^2+16)-25}{x(x+3)(\sqrt{x^2+16}+5)}\right)=\\\\\\\lim_{x\rightarrow-3}\frac{(x^2-9)}{x(x+3)(\sqrt{x^2+16}+5)}=\\\\\\\lim_{x\rightarrow-3}\frac{(x+3)(x-3)}{x(x+3)(\sqrt{x^2+16}+5)}=\\\\\\\lim_{x\rightarrow-3}\frac{(x-3)}{x(\sqrt{x^2+16}+5)}=$

$\frac{-3-3}{-3(\sqrt{9+16}+5)}=\\\\\\\frac{-6}{-3\cdot10}=\\\\\\\frac{-6^{\div(-6}}{-30^{\div(-6}}=\\\\\\\boxed{\frac{1}{5}}}$