To evaluate the limit

$\lim_{r \to 0} \frac{\sin\left(r^3\cos^3(\theta) + r^2\sin^2(\theta)\right)}{r^2},$

we need to consider the behavior of the function as $r \to 0$. Let's break down the expression and analyze it step by step.

Step 1: Approximation of the sine function for small arguments

Recall that for small $x$, $\sin(x)$ can be approximated by $x$, i.e., $\sin(x) \approx x \quad \text{as} \quad x \to 0.$ Thus, when $r$ is close to 0, we can approximate: $\sin\left(r^3\cos^3(\theta) + r^2\sin^2(\theta)\right) \approx r^3\cos^3(\theta) + r^2\sin^2(\theta).$

Step 2: Simplifying the limit expression

Now substitute the approximation into the original limit: $\lim_{r \to 0} \frac{r^3\cos^3(\theta) + r^2\sin^2(\theta)}{r^2}.$

We can now separate the terms in the numerator: $\lim_{r \to 0} \left( \frac{r^3\cos^3(\theta)}{r^2} + \frac{r^2\sin^2(\theta)}{r^2} \right).$ Simplifying each term: $= \lim_{r \to 0} \left( r\cos^3(\theta) + \sin^2(\theta) \right).$

Step 3: Evaluate the limit

As $r \to 0$, the term $r\cos^3(\theta)$ tends to 0. Therefore, the limit becomes: $\lim_{r \to 0} \sin^2(\theta).$

Thus, the limit is: $\boxed{\sin^2(\theta)}.$

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Do you want more details on any step, or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the general approximation for $\sin(x)$ near $x = 0$?
2. How does the small-angle approximation simplify trigonometric limits?
3. Can you explain why the higher-order terms vanish as $r \to 0$?
4. What would happen if we replaced $\sin(x)$ with $\cos(x)$ in the original problem?
5. How can we use Taylor series to refine the approximation in this limit?

Tip: When evaluating limits involving trigonometric functions, always check if small-angle approximations can simplify the expression, especially as the variable approaches zero.