Question

I need your help. Steps please. Thank you

Answer 1

$\lim\limits_{x\to 1}\dfrac{\sqrt{ax+b}-\sqrt{x+1}}{x^2-1}=4\sqrt{2}$

For the limit to be finite the numerator has to be $0$ : 

$\sqrt{a(1)+b}-\sqrt{1+1} = 2 \implies a+b = 2 \implies b = 2-a$

Plugging this in the earlier equation gives 

$\lim\limits_{x\to 1}\dfrac{\sqrt{ax+2-a}-\sqrt{x+1}}{x^2-1} = 4\sqrt{2}$

rationalize numerator on left hand side and get
$\lim\limits_{x\to 1}\dfrac{ax+2-a-x-1}{(x^2-1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{a(x-1)-1(x-1)}{(x^2-1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{(x-1)(a-1)}{(x-1)(x+1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{a-1}{(x+1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\dfrac{a-1}{(1+1)(\sqrt{a(1)+2-a}+\sqrt{1+1})} = 4\sqrt{2}\\\dfrac{a-1}{4\sqrt{2}} = 4\sqrt{2}$
You can solve $a$