We are tasked with evaluating the limit:

$\lim_{h \to 0} \frac{(1 + h)^2 - 1}{h}$

Step 1: Expand the expression $(1 + h)^2 - 1$

First, expand $(1 + h)^2$:

$(1 + h)^2 = 1^2 + 2 \cdot 1 \cdot h + h^2 = 1 + 2h + h^2$

Thus, the expression becomes:

$\frac{(1 + h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h}$

Simplify:

$\frac{2h + h^2}{h}$

Step 2: Factor out $h$

$\frac{2h + h^2}{h} = \frac{h(2 + h)}{h}$

Cancel out $h$ from the numerator and denominator (for $h \neq 0$):

$2 + h$

Step 3: Take the limit as $h \to 0$

Now, apply the limit:

$\lim_{h \to 0} (2 + h) = 2 + 0 = 2$

Final Answer:

$\lim_{h \to 0} \frac{(1 + h)^2 - 1}{h} = 2$

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Would you like any further explanation on this solution, or do you have any specific questions?

Additional Questions for You:

1. How does this type of limit relate to the definition of the derivative?
2. Can you solve a similar limit, such as $\lim_{h \to 0} \frac{(1 + h)^3 - 1}{h}$?
3. How would you compute the derivative of $f(x) = (1 + x)^2$ using limits?
4. How does the factorization step in solving limits help in simplifying expressions?
5. What is the geometric interpretation of limits in calculus?

Tip: When solving limits of this form, it's useful to first simplify the expression before directly substituting values, as direct substitution might lead to indeterminate forms like $\frac{0}{0}$.