Let's solve the given limit step by step:

We are asked to find the limit:

$\lim_{h \to 0} \frac{(h+3)^2 - 9}{h}$

Step 1: Expand the numerator

First, let's expand $(h+3)^2$: $(h+3)^2 = h^2 + 6h + 9$ Now, substitute this back into the original expression: $\frac{(h^2 + 6h + 9) - 9}{h} = \frac{h^2 + 6h + 9 - 9}{h} = \frac{h^2 + 6h}{h}$

Step 2: Simplify the expression

Now, factor the numerator: $\frac{h^2 + 6h}{h} = \frac{h(h + 6)}{h}$ Since $h \neq 0$ (we are approaching $h = 0$, not evaluating at it), we can cancel out $h$: $h + 6$

Step 3: Evaluate the limit

Now that we've simplified the expression to $h + 6$, we can evaluate the limit as $h \to 0$: $\lim_{h \to 0} (h + 6) = 0 + 6 = 6$

Conclusion:

The limit is $\boxed{6}$.

Do you need further clarification or have any questions?

Here are 5 related questions to explore:

1. What happens if we apply L'Hospital's rule in this case?
2. How do limits change when the indeterminate form is different, such as $\frac{\infty}{\infty}$?
3. What other algebraic techniques can be used to simplify complex limits?
4. What is the significance of limits in the definition of derivatives?
5. How does the cancellation of terms affect the outcome of limits?

Tip: Always check if a simple algebraic manipulation can resolve the limit before considering advanced techniques like L'Hospital's Rule.