We need to evaluate the limit of the expression $\frac{x-3}{1-\sqrt{4-x}}$ as $x$ approaches $3$.

1. Problem Description

We need to evaluate the limit of the expression $\frac{x-3}{1-\sqrt{4-x}}$ as $x$ approaches $3$.

2. Solution Steps

First, we observe that directly substituting $x=3$ into the expression results in the indeterminate form $\frac{0}{0}$.

Therefore, we need to manipulate the expression to resolve the indeterminate form.

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

So, we have

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

Using the difference of squares, $(a-b)(a+b) = a^2 - b^2$, we can simplify the denominator.

$(1-\sqrt{4-x})(1+\sqrt{4-x}) = 1^2 - (\sqrt{4-x})^2 = 1 - (4-x) = 1 - 4 + x = x - 3$

Now, the expression becomes

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

We can cancel the $(x-3)$ terms in the numerator and denominator:

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

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

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

3. Final Answer

The limit is

2.