To find the instantaneous rate of change of $f(x) = x^2 - 2$ at $x = 1$ using the limit definition, we'll use the formula for the derivative at a point:

$f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}$

Step-by-Step Solution:

1. Find $f(1 + h)$:

   Since $f(x) = x^2 - 2$, we substitute $1 + h$ into the function:

   $f(1 + h) = (1 + h)^2 - 2$

   Expand $(1 + h)^2$:

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

   So,

   $f(1 + h) = 1 + 2h + h^2 - 2 = h^2 + 2h - 1$

2. Find $f(1)$:

   Using the function $f(x) = x^2 - 2$, substitute $x = 1$:

   $f(1) = 1^2 - 2 = 1 - 2 = -1$

3. Set up the difference quotient:

   Substitute $f(1 + h)$ and $f(1)$ into the difference quotient formula:

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

4. Simplify the expression:

   Factor out $h$ from the numerator:

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

   Cancel the $h$ in the numerator and denominator (assuming $h \neq 0$):

   $h + 2$

5. Take the limit as $h \to 0$:

   Now, find the limit as $h \to 0$:

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

Conclusion:

The instantaneous rate of change of $f(x) = x^2 - 2$ at $x = 1$ is $2$.

---

Do you want more details, or have any other questions? Here are 5 related questions:

1. What does the limit definition of a derivative represent geometrically?
2. Can you use the same process to find the derivative at $x = 2$?
3. What is the general formula for the derivative of $f(x) = x^2 - 2$?
4. How would you find the instantaneous rate of change for a different function, like $f(x) = x^3 - 2x$?
5. Can you explain why the $h$ term can be canceled in the simplification step?

Tip: The derivative of a function at a point gives the slope of the tangent line to the curve at that point.