Analyzing Pointwise Convergence of Function Sequence $f_n(x) = x^n$

The question asks us to determine the convergence behavior of the sequence of functions $\{f_n(x) = x^n\}$ defined on the interval $[0,1]$. We need to compare this sequence's limit to the given function $f(x)$.

Function Definitions

We are given:

• The sequence of functions: $f_n(x) = x^n$, where $n = 1, 2, \dots$ and $x \in [0,1]$.
• The limit function: $f(x) = \begin{cases} 0, & \forall x \in [0,1) \\ 1, & x = 1 \end{cases}$

Determining Pointwise Convergence

Pointwise convergence means that for each individual value of $x$ in the domain, the sequence of numbers $\{f_n(x)\}$ converges to $f(x)$. We examine the limit of $f_n(x)$ as $n \to \infty$ for different values of $x$ in $[0,1]$.

• For $x$ in the interval $[0, 1)$:

  If $x$ is strictly less than 1 (e.g., $x = 0.5$), raising $x$ to higher powers makes the result smaller. As $n$ becomes very large, $x^n$ approaches 0.

  For example, $(0.5)^1 = 0.5$, $(0.5)^2 = 0.25$, $(0.5)^{10} \approx 0.00097$.

  Therefore, for $x \in [0,1)$, we have $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} x^n = 0$.

• For $x = 1$:

  If $x = 1$, then $f_n(1) = 1^n = 1$ for any positive integer $n$. The sequence is constantly 1.

  Therefore, for $x = 1$, we have $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} 1^n = 1$.

By combining these results, the pointwise limit function, let's call it $L(x)$, is:

This limit function $L(x)$ is exactly the same as the function $f(x)$ given in the problem. This confirms that the sequence $\{f_n\}$ converges pointwise to $f$.

Evaluating Uniform Convergence (for context)

Uniform convergence requires the maximum difference between $f_n(x)$ and $f(x)$ over the entire interval $[0,1]$ to approach 0 as $n \to \infty$.

We look at the difference $|f_n(x) - f(x)|$.

• For $x \in [0,1)$, $|f_n(x) - f(x)| = |x^n - 0| = x^n$.
• For $x=1$, $|f_n(x) - f(x)| = |1^n - 1| = 0$.

The maximum value of $|f_n(x) - f(x)|$ occurs as $x$ approaches 1. The supremum of $x^n$ on $[0,1)$ is 1.

So, $\sup_{x \in [0,1]} |f_n(x) - f(x)| = 1$ for all $n$.

Since $\lim_{n \to \infty} 1 = 1 \neq 0$, the convergence is not uniform.

Conclusion

The sequence of functions $\{f_n(x) = x^n\}$ converges pointwise to $f(x)$ on the interval $[0,1]$. This aligns with the first option provided.