To find the limit as $h$ approaches 0 for the expression $\frac{\frac{1}{(x+h)^3} - \frac{1}{x^3}}{h}$, we can use the definition of the derivative. This limit represents the derivative of $f(x) = \frac{1}{x^3}$ with respect to $x$.

Step-by-Step Solution:

1. Rewrite the limit expression: $\lim_{h \to 0} \frac{\frac{1}{(x+h)^3} - \frac{1}{x^3}}{h}$

2. Combine the fractions in the numerator: $\frac{1}{(x+h)^3} - \frac{1}{x^3} = \frac{x^3 - (x+h)^3}{x^3(x+h)^3}$ So, the limit becomes: $\lim_{h \to 0} \frac{\frac{x^3 - (x+h)^3}{x^3(x+h)^3}}{h} = \lim_{h \to 0} \frac{x^3 - (x+h)^3}{h \cdot x^3(x+h)^3}$

3. Expand the expression $(x+h)^3$: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ Therefore: $x^3 - (x+h)^3 = x^3 - \left(x^3 + 3x^2h + 3xh^2 + h^3\right) = -3x^2h - 3xh^2 - h^3$

4. Simplify the numerator: $\lim_{h \to 0} \frac{-3x^2h - 3xh^2 - h^3}{h \cdot x^3(x+h)^3}$ Factor $h$ out of the numerator: $\lim_{h \to 0} \frac{h(-3x^2 - 3xh - h^2)}{h \cdot x^3(x+h)^3}$ Cancel $h$ from the numerator and denominator: $\lim_{h \to 0} \frac{-3x^2 - 3xh - h^2}{x^3(x+h)^3}$

5. Take the limit as $h$ approaches 0: $\lim_{h \to 0} \frac{-3x^2 - 3xh - h^2}{x^3(x+h)^3} = \frac{-3x^2}{x^3 \cdot x^3} = \frac{-3x^2}{x^6} = -\frac{3}{x^4}$

Final Answer:

$\lim_{h \to 0} \frac{\frac{1}{(x+h)^3} - \frac{1}{x^3}}{h} = -\frac{3}{x^4}$

Would you like more details on any of these steps, or do you have any questions?

Related Questions:

1. What is the derivative of $f(x) = \frac{1}{x^n}$ in general?
2. How does this approach change if the power of $x$ is different, say $x^2$ instead of $x^3$?
3. What are the rules of differentiation that apply to rational functions?
4. How do we use limits to derive the derivative of trigonometric functions?
5. What happens if you use an alternative method like the quotient rule for this problem?

Tip:

When simplifying limits that involve small changes like $h$ approaching zero, expanding terms using binomial expansion can be very helpful in identifying leading terms for simplification.