We are asked to find the limit:

$\lim_{h \to 0} \frac{f(5+h) - f(5)}{h}$

where $f(x) = 6x - 1$.

Step 1: Substitute $f(x)$ into the expression

We have $f(x) = 6x - 1$, so:

• $f(5+h) = 6(5+h) - 1 = 30 + 6h - 1 = 29 + 6h$,
• $f(5) = 6(5) - 1 = 30 - 1 = 29$.

Step 2: Compute the difference

Now, compute $f(5+h) - f(5)$: $f(5+h) - f(5) = (29 + 6h) - 29 = 6h.$

Step 3: Substitute into the limit expression

Substitute $f(5+h) - f(5) = 6h$ into the original limit: $\lim_{h \to 0} \frac{6h}{h}.$

Step 4: Simplify the expression

Simplify the expression: $\frac{6h}{h} = 6.$

Step 5: Evaluate the limit

As $h \to 0$, the simplified expression is simply: $\lim_{h \to 0} 6 = 6.$

Thus, the solution to the limit is $\boxed{6}$.

Would you like further details or have any questions? Here are some related questions to extend the topic:

1. What is the geometric interpretation of this limit?
2. How does the derivative relate to this limit problem?
3. What is the derivative of $f(x) = 6x - 1$ in general?
4. Can the limit be solved using L'Hopital's Rule?
5. How does this limit change if $f(x)$ were a quadratic function?

Tip: The limit we just computed is the derivative of the function $f(x) = 6x - 1$ at $x = 5$.