The limit you are referring to is the definition of the derivative of a function $f(x)$ at a point $x$. In this case, we are tasked with evaluating the derivative of $f(x) = x^2$ at $x = -1$.

The general form of the derivative definition is:

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

Step-by-Step Solution:

1. Start with the function: $f(x) = x^2$

2. Apply the limit definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{(f(x+h) - f(x))}{h}$

3. Substitute the given function $f(x) = x^2$ into the expression: $f'(x) = \lim_{h \to 0} \frac{((x+h)^2 - x^2)}{h}$

4. Expand $(x+h)^2$: $(x+h)^2 = x^2 + 2xh + h^2$ So the expression becomes: $f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$

5. Simplify by canceling terms: $f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$ Factor $h$ out of the numerator: $f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h}$

6. Cancel $h$ from the numerator and denominator: $f'(x) = \lim_{h \to 0} (2x + h)$

7. Evaluate the limit as $h \to 0$: $f'(x) = 2x$

8. Finally, substitute $x = -1$: $f'(-1) = 2(-1) = -2$

Final Answer:

The derivative of $f(x) = x^2$ at $x = -1$ is $f'(-1) = -2$.

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Would you like further details on this process or any part of the derivative definition?

Here are 5 additional questions for deeper understanding:

1. How is the derivative related to the slope of the tangent line?
2. What does the derivative represent geometrically?
3. How would you differentiate $f(x) = x^3$ using the same limit definition?
4. What happens if you evaluate the derivative at $x = 1$ for $f(x) = x^2$?
5. How can the power rule be derived from the limit definition of a derivative?

Tip: Always expand and simplify algebraic expressions when working with limits involving derivatives!