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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 \( x = 5 \).

Find lim(h→0) [(f(5+h) - f(5))/h] if f(x) = 6x - 1. Simplify your answer.

Learn how to find the limit of (f(5+h) - f(5))/h as h approaches 0 for the function f(x) = 6x - 1. This step-by-step solution covers key calculus concepts, including the limit definition of the derivative. Suitable for high school students studying Calculus.

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