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We are tasked with finding the limit of the difference quotient:

\[
\lim_{{h \to 0}} \frac{{f(4+h) - f(4)}}{h}
\]
where \( f(x) = |5x - 1| \).

### Step 1: Compute \( f(4) \)

Substitute \( x = 4 \) into the given function \( f(x) \):
\[
f(4) = |5(4) - 1| = |20 - 1| = |19| = 19
\]

### Step 2: Compute \( f(4+h) \)

We need to evaluate \( f(4 + h) \):
\[
f(4+h) = |5(4+h) - 1| = |20 + 5h - 1| = |19 + 5h|
\]

### Step 3: Behavior of \( f(4+h) \)

Now, we consider two cases for the absolute value \( |19 + 5h| \):

- **When \( h \geq 0 \):** \( f(4+h) = 19 + 5h \)
- **When \( h < 0 \):** \( f(4+h) = -(19 + 5h) = -19 - 5h \)

### Step 4: Left-Hand and Right-Hand Limits

#### Right-hand limit \( h \to 0^+ \):
For \( h > 0 \), we use \( f(4+h) = 19 + 5h \):
\[
\frac{{f(4+h) - f(4)}}{h} = \frac{{(19 + 5h) - 19}}{h} = \frac{{5h}}{h} = 5
\]

Thus, as \( h \to 0^+ \), the limit is 5.

#### Left-hand limit \( h \to 0^- \):
For \( h < 0 \), we use \( f(4+h) = -19 - 5h \):
\[
\frac{{f(4+h) - f(4)}}{h} = \frac{{(-19 - 5h) - 19}}{h} = \frac{{-38 - 5h}}{h} = \frac{{-5h}}{h} = -5
\]

Thus, as \( h \to 0^- \), the limit is \(-5\).

### Step 5: Conclusion

The left-hand and right-hand limits are not equal. Therefore, the limit does not exist. The limit of the difference quotient is undefined.

Would you like more details or have any questions?

Here are 5 related questions for further understanding:
1. What is the geometric interpretation of a derivative when a limit doesn't exist?
2. Can a function with an absolute value have a smooth derivative everywhere?
3. How do piecewise functions affect the calculation of limits?
4. What role does the sign of \( h \) play in limits involving absolute value functions?
5. What conditions must be met for the derivative to exist at a point of non-differentiability?

**Tip**: When dealing with absolute value functions, always break the function into cases to understand its behavior near critical points like zero.

Find lim (h→0) [f(4+h) - f(4)] / h if f(x) = |5x - 1|.

Learn how to find the limit of a difference quotient for the function f(x) = |5x - 1| at x = 4. This problem involves evaluating one-sided limits and understanding the behavior of absolute value functions. Suitable for Grades 11-12 students.

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