We are tasked with finding the limit:

\[
\lim_{{x \to 4}} \frac{4 - \sqrt{5x - 4}}{4 - x}
\]

### Step 1: Substitution to check for direct evaluation
If we directly substitute \( x = 4 \) into the expression:

\[
\frac{4 - \sqrt{5(4) - 4}}{4 - 4} = \frac{4 - \sqrt{20 - 4}}{0} = \frac{4 - \sqrt{16}}{0} = \frac{4 - 4}{0} = \frac{0}{0}
\]

This results in an indeterminate form \( \frac{0}{0} \), so we need to apply algebraic techniques to resolve the limit.

### Step 2: Rationalize the numerator
We rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator:

\[
\text{Conjugate of } \left(4 - \sqrt{5x - 4}\right) = \left(4 + \sqrt{5x - 4}\right)
\]

Multiply the entire fraction by:

\[
\frac{4 + \sqrt{5x - 4}}{4 + \sqrt{5x - 4}}
\]

This gives us:

\[
\lim_{{x \to 4}} \frac{(4 - \sqrt{5x - 4})(4 + \sqrt{5x - 4})}{(4 - x)(4 + \sqrt{5x - 4})}
\]

### Step 3: Simplify the expression
Using the difference of squares formula:

\[
(a - b)(a + b) = a^2 - b^2
\]

we get:

\[
\lim_{{x \to 4}} \frac{16 - (5x - 4)}{(4 - x)(4 + \sqrt{5x - 4})}
\]

Simplifying the numerator:

\[
16 - (5x - 4) = 16 - 5x + 4 = 20 - 5x
\]

Thus, the expression becomes:

\[
\lim_{{x \to 4}} \frac{20 - 5x}{(4 - x)(4 + \sqrt{5x - 4})}
\]

### Step 4: Factor and cancel
Notice that \( 20 - 5x \) can be factored as:

\[
20 - 5x = 5(4 - x)
\]

Substituting this in, we get:

\[
\lim_{{x \to 4}} \frac{5(4 - x)}{(4 - x)(4 + \sqrt{5x - 4})}
\]

Now, cancel the common term \( (4 - x) \) from the numerator and denominator:

\[
\lim_{{x \to 4}} \frac{5}{4 + \sqrt{5x - 4}}
\]

### Step 5: Substitute \( x = 4 \)
Now that we've simplified the expression, we can directly substitute \( x = 4 \):

\[
\frac{5}{4 + \sqrt{5(4) - 4}} = \frac{5}{4 + \sqrt{20 - 4}} = \frac{5}{4 + \sqrt{16}} = \frac{5}{4 + 4} = \frac{5}{8}
\]

Thus, the limit is:

\[
\boxed{\frac{5}{8}}
\]

Would you like further clarification on any step?

### Related Questions:
1. How do you rationalize an expression in calculus?
2. What is the difference of squares formula and when is it used?
3. How do you solve limits that result in the indeterminate form \( \frac{0}{0} \)?
4. What are some techniques to handle limits with radicals?
5. How can factoring help in simplifying limits?

#### Tip:
Always check for direct substitution in limits before applying any algebraic manipulations to save time.

Explore how to solve the limit lim(x→4) (4 - √(5x - 4))/(4 - x) with a step-by-step solution. Key calculus concepts include indeterminate forms, rationalization, and the difference of squares theorem. Perfect for college calculus or advanced high school students.

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