Question

lim
X→4
$\frac{ \sqrt{2( {x} ^{2} - 8) + x} }{x + 4}$
​

Answer 1

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

$=\dfrac{\sqrt{20}}{8}=\dfrac{2\sqrt{5}}{8}=\boxed{\dfrac{\sqrt{5}}{4}}$