Question

LIMITES

√2($x^{2}$ -8) +x / x+4 com x tendendo a -4

obs: a raiz no numerador inclui 2(x^2-8)

Answer 1

$\lim_{x \to \ -4 \frac{ \sqrt{2(x^2-8)}+x }{x+4}$
$=\frac{ \sqrt{2(-4^2+8)} -4}{-4+4}$
$=\frac{ \sqrt{2(16-8)} -4}{0}$
$=\frac{ \sqrt{16} -4}{0}=\frac{4-4}{0} = \frac{0}{0} (SI)$

$\lim_{x \to \ -4 \frac{ \sqrt{2(x^2-8)}+x }{x+4}.\frac{ \sqrt{2(x^2-8)}-x }{ \sqrt{2(x^2-8)}-x } }$

$(a+b)(a-b)=a^2-b^2$
$( \sqrt{2(x^2-8)}-x } )( \sqrt{2(x^2-8)}+x } )=(\sqrt{2(x^2-8)})^2-x^2$
$( \sqrt{2(x^2-8)}-x } )( \sqrt{2(x^2-8)}+x } )=2(x^2-8)-x$

$\lim_{x \to \ -4 \frac{2(x^2-8)-x}{x+4 \sqrt{2(x^2-8)-x} }$
$\lim_{x \to \ -4 \frac{2(x^2-8)}{x+4 \sqrt{2(x^2-8)} }$
$=\frac{2(-4^2-8)}{-4+4 \sqrt{2(-4^2-8)} }$
$=\frac{2(16-8)}{ \sqrt{2(16-8)} }$
$=\frac{16}{ \sqrt{16} }=\frac{16}{4 }= \frac{4}{1}$