To solve the limit:

$\lim_{x \to \infty} \frac{(-3)^x}{x^2 3^x}$

Let's break this expression down step by step.

Step 1: Simplifying the expression

First, rewrite the expression to separate the powers of $3$:

$\frac{(-3)^x}{x^2 3^x} = \frac{(-1)^x 3^x}{x^2 3^x}$

Notice that $3^x$ appears in both the numerator and denominator, so we can cancel them:

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

Now the expression is:

$\frac{(-1)^x}{x^2}$

Step 2: Analyzing the behavior of the expression

As $x \to \infty$, we look at the two components of the fraction:

1. $(-1)^x$ alternates between $-1$ and $1$, depending on whether $x$ is odd or even.
2. $x^2$ grows without bound as $x \to \infty$.

So, the overall expression $\frac{(-1)^x}{x^2}$ becomes very small in magnitude because $x^2$ grows large, while $(-1)^x$ only oscillates between $-1$ and $1$.

Step 3: Taking the limit

Since $x^2 \to \infty$, the fraction $\frac{(-1)^x}{x^2}$ tends to $0$ as $x \to \infty$.

Thus:

$\lim_{x \to \infty} \frac{(-1)^x}{x^2} = 0$

Conclusion:

The limit of the given expression is:

$\boxed{0}$

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Would you like a more detailed breakdown of any part of this solution? Here are some related questions to explore further:

1. What happens if we replace $(-1)^x$ with $1^x$ in the expression?
2. How can we evaluate limits involving alternating series like this one?
3. What if the exponent in $(-3)^x$ was a continuous function of $x$?
4. How does the growth of polynomial functions compare to exponential functions in limits?
5. How would the limit change if the denominator had a different power, such as $x^3$?

Tip: When dealing with limits where the numerator oscillates (like $(-1)^x$) and the denominator grows without bound, you can often deduce that the limit is zero by recognizing that the growth of the denominator "dominates" the oscillations.