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To solve this problem, we need to evaluate the limit:

\[
\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}
\]

This expression represents the derivative of \( f(x) = 7x - 3 \) at \( x = 2 \).

### Step 1: Find \( f(2) \)
Using the function \( f(x) = 7x - 3 \):
\[
f(2) = 7 \cdot 2 - 3 = 14 - 3 = 11
\]

### Step 2: Find \( f(2 + h) \)
Substitute \( x = 2 + h \) into \( f(x) = 7x - 3 \):
\[
f(2 + h) = 7(2 + h) - 3 = 14 + 7h - 3 = 11 + 7h
\]

### Step 3: Set up the limit expression
Now substitute \( f(2 + h) \) and \( f(2) \) into the difference quotient:
\[
\frac{f(2+h) - f(2)}{h} = \frac{(11 + 7h) - 11}{h} = \frac{7h}{h}
\]

### Step 4: Simplify the expression
\[
\frac{7h}{h} = 7 \quad \text{(for \( h \neq 0 \))}
\]

### Step 5: Take the limit as \( h \to 0 \)
Since there are no \( h \)-terms left, the limit is:
\[
\lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = 7
\]

### Final Answer
\[
\boxed{7}
\]

Would you like more details, or do you have any questions?

### Related Questions
1. How would the answer change if \( f(x) = 7x + 3 \) instead of \( 7x - 3 \)?
2. Can you find the derivative of \( f(x) = ax + b \) at any point \( x = c \)?
3. What does the derivative of a linear function represent geometrically?
4. How would you apply the limit definition of a derivative to a quadratic function, like \( f(x) = x^2 \)?
5. How can you compute the derivative of \( f(x) = 7x - 3 \) using the power rule instead?

### Tip
When dealing with linear functions, their derivatives are constants because they represent the slope of the line across all values of \( x \).

Find the following limit for f(x) = 7x - 3: lim(h→0) [f(2+h) - f(2)] / h

Learn how to find the limit of the expression lim(h→0) [f(2+h) - f(2)] / h for the function f(x) = 7x - 3 using the limit definition of a derivative. This example demonstrates calculating the derivative of a linear function.

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