We are tasked with evaluating the limit:

$\lim_{x \to 1} \frac{\sqrt{2 - x} - 1}{3 - \sqrt{4x + 5}}$

Step 1: Substitution of $x = 1$

First, let's substitute $x = 1$ into the expression to see if we get an indeterminate form or a specific value:

$\frac{\sqrt{2 - 1} - 1}{3 - \sqrt{4(1) + 5}} = \frac{\sqrt{1} - 1}{3 - \sqrt{9}} = \frac{1 - 1}{3 - 3} = \frac{0}{0}$

Since this results in the indeterminate form $\frac{0}{0}$, we need to apply L'Hopital's Rule or other analytical methods.

Step 2: Apply L'Hopital's Rule

L'Hopital's Rule states that if the limit results in $\frac{0}{0}$, we can differentiate the numerator and the denominator separately and then take the limit again. Let's proceed by differentiating the numerator and denominator.

Numerator: $f(x) = \sqrt{2 - x} - 1$

The derivative of the numerator $f(x) = \sqrt{2 - x} - 1$ is:

$f'(x) = \frac{d}{dx} \left( \sqrt{2 - x} \right) = \frac{-1}{2\sqrt{2 - x}}$

Denominator: $g(x) = 3 - \sqrt{4x + 5}$

The derivative of the denominator $g(x) = 3 - \sqrt{4x + 5}$ is:

[ g'(x) = \frac{d}{dx} \left( 3 - \sqrt{4x + 5} \right) = - \frac{4}{2\sqrt{4x + 5}} = \frac{-2}{\sqrt{4x + 5}}