The problem implies Xn is a binomial random variable, Xn Bin(n, p_n), as this is the context for such limit theorems. We are given two conditions:

The problem implies $X_n$ is a binomial random variable, $X_n \sim Bin(n, p_n)$, as this is the context for such limit theorems. We are given two conditions:

1. $p_n \to 0$ as $n \to \infty$. This means the probability of success for each trial becomes very small.
2. $np_n \to \infty$ as $n \to \infty$. This means the expected number of successes grows infinitely large.

• A. Binomial: $X_n$ is already a binomial random variable. The question asks for an approximation, implying a different distribution that $X_n$ converges to under these conditions, not its original form.
• B. Poisson: The Poisson approximation to the binomial distribution is valid when $n \to \infty$, $p_n \to 0$, and $np_n \to \lambda$, where $\lambda$ is a finite positive constant. In this problem, $np_n \to \infty$, so the Poisson approximation is not appropriate.
• C. Bernoulli: A Bernoulli distribution describes a single trial ($n=1$). It is not an asymptotic approximation for a sequence of $n$ trials as $n \to \infty$.
• D. Normal: The Normal approximation to the binomial distribution (also known as the De Moivre-Laplace theorem, a special case of the Central Limit Theorem) states that if $X_n \sim Bin(n, p_n)$, then for large $n$, $X_n$ can be approximated by a normal distribution $N(\mu, \sigma^2)$ where $\mu = np_n$ and $\sigma^2 = np_n(1-p_n)$. The conditions for this approximation to be good are typically that both $np_n$ and $n(1-p_n)$ are sufficiently large (e.g., $\ge 5$ or 10). Given $np_n \to \infty$, the mean is growing large. Given $p_n \to 0$, it implies $1-p_n \to 1$. Since $np_n \to \infty$ and $p_n \to 0$, $n$ must also tend to infinity. Therefore, $n(1-p_n) \approx n \to \infty$. Since both $np_n \to \infty$ and $n(1-p_n) \to \infty$, the conditions for the normal approximation are met. The standardized variable $\frac{X_n - np_n}{\sqrt{np_n(1-p_n)}}$ converges in distribution to a standard normal distribution.

Based on the conditions $p_n \to 0$ and $np_n \to \infty$, the binomial distribution $X_n$ is best approximated by a Normal distribution.

The final answer is $\boxed{\text{D}}$