To find $f'(2)$ using the limit definition of the derivative, we apply the formula:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

In this case, $f(x) = x^2 + 5x + 1$, and we are asked to find $f'(2)$. So we substitute $x = 2$ into the limit definition:

$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

Step 1: Calculate $f(2+h)$

$f(2+h) = (2+h)^2 + 5(2+h) + 1$ Expanding this: $f(2+h) = (4 + 4h + h^2) + (10 + 5h) + 1$ $f(2+h) = h^2 + 9h + 15$

Step 2: Calculate $f(2)$

$f(2) = 2^2 + 5(2) + 1 = 4 + 10 + 1 = 15$

Step 3: Substitute into the limit formula

Now substitute these values into the limit definition: $f'(2) = \lim_{h \to 0} \frac{(h^2 + 9h + 15) - 15}{h}$ Simplify the numerator: $f'(2) = \lim_{h \to 0} \frac{h^2 + 9h}{h}$ Factor out $h$ from the numerator: $f'(2) = \lim_{h \to 0} \frac{h(h + 9)}{h}$ Cancel out $h$ (for $h \neq 0$): $f'(2) = \lim_{h \to 0} (h + 9)$

Step 4: Evaluate the limit

As $h \to 0$, we get: $f'(2) = 9$

Thus, $f'(2) = 9$.

Would you like further details on any part of this process or have any questions?

Here are five related questions you might want to explore:

1. How do you find the derivative of $f(x) = x^2 + 5x + 1$ using standard differentiation rules?
2. Can you explain the physical meaning of the derivative at a point?
3. How does the limit definition of the derivative differ from simply applying differentiation rules?
4. What happens if we try to calculate $f'(x)$ at a point where the function is not differentiable?
5. How would the calculation change if the function had a higher degree, such as $f(x) = x^3 + 5x + 1$?

Tip: The limit definition of the derivative helps you understand differentiation from first principles, showing the derivative as the slope of the tangent line at a specific point.