Để giới hạn hữu hạn: $a=1$
$\lim\limits_{x\to -\infty}(\sqrt{x^2+4x}+x+b)$
$=\lim\limits_{x\to -\infty}\dfrac{x^2+4x-(x^2+2bx+b^2)}{\sqrt{x^2+4x}-(x+b)}$
$=\lim\limits_{x\to -\infty}\dfrac{(4-2b)x-b^2 }{\sqrt{x^2\Big( 1+\dfrac{4}{x}\Big)}-x-b}$
$=\lim\limits_{x\to -\infty}\dfrac{ 4-2b-\dfrac{b^2}{x}}{ -\sqrt{1+\dfrac{4}{x}} -1-\dfrac{b}{x}}$
$=\dfrac{4-2b}{-1-1}=10$
$\to b=12$
Vậy $(a;b)=(1;12)$
