We need to evaluate the limit: $\lim_{x\to 3} \frac{x-3}{1 - \sqrt{4-x}}$.

1. Problem Description

We need to evaluate the limit:

$\lim_{x\to 3} \frac{x-3}{1 - \sqrt{4-x}}$.

2. Solution Steps

First, we notice that if we substitute $x=3$ directly into the expression, we get $\frac{3-3}{1-\sqrt{4-3}} = \frac{0}{1-1} = \frac{0}{0}$, which is an indeterminate form. Therefore, we need to manipulate the expression to evaluate the limit.

We can multiply the numerator and denominator by the conjugate of the denominator, which is $1+\sqrt{4-x}$.

$\lim_{x\to 3} \frac{x-3}{1-\sqrt{4-x}} = \lim_{x\to 3} \frac{(x-3)(1+\sqrt{4-x})}{(1-\sqrt{4-x})(1+\sqrt{4-x})}$

$= \lim_{x\to 3} \frac{(x-3)(1+\sqrt{4-x})}{1 - (4-x)}$

$= \lim_{x\to 3} \frac{(x-3)(1+\sqrt{4-x})}{1-4+x}$

$= \lim_{x\to 3} \frac{(x-3)(1+\sqrt{4-x})}{x-3}$

Now, we can cancel the $(x-3)$ terms in the numerator and denominator, since we are considering the limit as $x$ approaches $3$, but $x$ is not equal to $3$.

$= \lim_{x\to 3} (1+\sqrt{4-x})$

Now, we can substitute $x=3$ into the simplified expression:

$1 + \sqrt{4-3} = 1 + \sqrt{1} = 1+1 = 2$

3. Final Answer

The final answer is 2.