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 $x$.