To determine if the given answers (a) and (c) are correct and check if there are any other options that also qualify, let's analyze each sequence for pointwise convergence on their respective domains.

1. Option (a): $f_n(x) = \frac{nx}{1 + n^2 x^2}$ on $D = [0, \infty)$

   For fixed $x$:

   • As $n \to \infty$, if $x = 0$, $f_n(0) = 0$.
   • If $x > 0$, $n^2 x^2$ grows faster than $nx$, so $f_n(x) \to 0$.

   Therefore, $f_n(x) \to 0$ for all $x \in [0, \infty)$. This sequence converges pointwise to the function $f(x) = 0$ on $[0, \infty)$.

2. Option (c): $f_n(x) = x^n$ on $D = [0, 1]$

   For fixed $x$:

   • If $x = 1$, $f_n(1) = 1^n = 1$.
   • If $0 \leq x < 1$, as $n \to \infty$, $x^n \to 0$ because $x^n$ gets smaller for values of $x$ in $[0, 1)$.

   Therefore, $f_n(x) \to 0$ for $x \in [0, 1)$ and $f_n(x) \to 1$ for $x = 1$. The sequence $f_n(x)$ converges pointwise to:

   0, & x \in [0, 1) \\ 1, & x = 1 \end{cases}$$ So, option (c) also converges pointwise on $$[0, 1]$$.

3. Option (b): $f_n(x) = n^2 x (1 - x)^n$ on $D = [0, 1]$

   For fixed $x$:

   • If $x = 0$, $f_n(0) = 0$.
   • For $x \in (0, 1]$, as $n \to \infty$, $(1 - x)^n \to 0$ since $(1 - x) < 1$. However, $n^2 x$ grows, and it is not clear that $f_n(x)$ converges pointwise to a single function over the entire interval $D = [0, 1]$.

   This sequence does not converge pointwise on $D = [0, 1]$.

4. Option (d): $f_n(x) = (1 + x^2)^n$ on $D = [0, 1]$

   For fixed $x$:

   • If $x \in [0, 1)$, $1 + x^2 > 1$, so $(1 + x^2)^n \to \infty$ as $n \to \infty$.
   • At $x = 1$, $(1 + 1^2)^n = 2^n \to \infty$.

   Thus, $f_n(x)$ diverges for all $x \in [0, 1]$.

Conclusion:

The correct options for pointwise convergence are:

• (a) $f_n(x) = \frac{nx}{1 + n^2 x^2}$ on $D = [0, \infty)$.
• (c) $f_n(x) = x^n$ on $D = [0, 1]$.

Therefore, the answers (a) and (c) are correct, and there are no other options that satisfy pointwise convergence.

Would you like a detailed explanation of any specific option, or have further questions?

Here are some related questions to deepen understanding:

1. What is the difference between pointwise convergence and uniform convergence?
2. How does pointwise convergence differ when considering functions on different domains?
3. Why does $x^n \to 0$ for $x \in [0, 1)$ but $x^n \to 1$ when $x = 1$?
4. What is the importance of pointwise convergence in functional analysis?
5. How can we test for uniform convergence on a given interval?

Tip: Pointwise convergence only ensures convergence at each point individually; it does not imply uniform convergence across the domain.