1. **Stating the problem:** We have a sequence of functions defined for $n=1,2,3,\ldots$ as $$f_n(x) = \frac{n^2 x^3}{1 + 2n^{2n^2}}$$ for every real number $x$. We want to find the function $$f(x) = \lim_{n \to \infty} f_n(x)$$ for each $x \in \mathbb{R}$. 2. **Understanding the limit:** The limit function $f(x)$ is the pointwise limit of the sequence $f_n(x)$ as $n$ approaches infinity. 3. **Analyzing the denominator:** The denominator is $$1 + 2n^{2n^2}.$$ As $n$ grows large, $n^{2n^2}$ grows extremely fast (much faster than any polynomial in $n$). 4. **Analyzing the numerator:** The numerator is $$n^2 x^3,$$ which grows polynomially in $n$. 5. **Comparing growth rates:** Since the denominator grows much faster than the numerator, the fraction $$\frac{n^2 x^3}{1 + 2n^{2n^2}}$$ tends to zero as $n \to \infty$ for every fixed $x$. 6. **Conclusion:** Therefore, $$f(x) = \lim_{n \to \infty} f_n(x) = 0$$ for all real numbers $x$. **Final answer:** $$\boxed{f(x) = 0 \text{ for all } x \in \mathbb{R}}.$$