We are asked to find the values of constants $a$ and $b$ such that $\lim_{x \to 1} \frac{a\sqrt{x+1} - b}{x-1} = \sqrt{2}$.

1. Problem Description

We are asked to find the values of constants $a$ and $b$ such that

$\lim_{x \to 1} \frac{a\sqrt{x+1} - b}{x-1} = \sqrt{2}$.

2. Solution Steps

Since the limit exists and is equal to $\sqrt{2}$, the numerator must approach 0 as $x$ approaches

1. This is because if the numerator does not approach zero while the denominator approaches zero, the limit will be infinite.

Thus,

$a\sqrt{1+1} - b = 0$

$a\sqrt{2} - b = 0$

$b = a\sqrt{2}$

Now, substitute $b = a\sqrt{2}$ into the limit:

$\lim_{x \to 1} \frac{a\sqrt{x+1} - a\sqrt{2}}{x-1} = \sqrt{2}$

$a\lim_{x \to 1} \frac{\sqrt{x+1} - \sqrt{2}}{x-1} = \sqrt{2}$

Multiply the numerator and denominator by the conjugate of the numerator:

$a\lim_{x \to 1} \frac{(\sqrt{x+1} - \sqrt{2})(\sqrt{x+1} + \sqrt{2})}{(x-1)(\sqrt{x+1} + \sqrt{2})} = \sqrt{2}$

$a\lim_{x \to 1} \frac{(x+1) - 2}{(x-1)(\sqrt{x+1} + \sqrt{2})} = \sqrt{2}$

$a\lim_{x \to 1} \frac{x-1}{(x-1)(\sqrt{x+1} + \sqrt{2})} = \sqrt{2}$

$a\lim_{x \to 1} \frac{1}{\sqrt{x+1} + \sqrt{2}} = \sqrt{2}$

$a\frac{1}{\sqrt{1+1} + \sqrt{2}} = \sqrt{2}$

$a\frac{1}{\sqrt{2} + \sqrt{2}} = \sqrt{2}$

$a\frac{1}{2\sqrt{2}} = \sqrt{2}$

$a = \sqrt{2}(2\sqrt{2}) = 2(2) = 4$

Since $b = a\sqrt{2}$, we have $b = 4\sqrt{2}$.

3. Final Answer

$a = 4$

$b = 4\sqrt{2}$