Solution:

We are tasked to find the pointwise limit of the sequence of functions $f_n(x) = x^n$ for $n \in \mathbb{N}$ on the closed interval $[0, 1]$, and to determine whether the convergence is uniform.

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Step 1: Pointwise Limit

For each fixed $x \in [0, 1]$, consider the sequence of values $f_n(x) = x^n$. We analyze the behavior of $x^n$ as $n \to \infty$.

1. Case 1: $x = 1$ f_n(1) = 1^n = 1 \quad \text{(for all n)}. So, $\lim_{n \to \infty} f_n(1) = 1$.

2. Case 2: $0 < x < 1$ For $0 < x < 1$, $x^n$ becomes smaller and smaller as $n \to \infty$ because $x < 1$ and repeated multiplication by $x$ results in values approaching $0$. Hence: $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} x^n = 0.$

3. Case 3: $x = 0$ f_n(0) = 0^n = 0 \quad \text{(for all n)}. So, $\lim_{n \to \infty} f_n(0) = 0$.

Conclusion (Pointwise Limit):

The pointwise limit function $f(x)$ is:

\begin{cases} 1 & \text{if } x = 1, \\ 0 & \text{if } 0 \leq x < 1. \end{cases}$$ --- #### **Step 2: Uniform Convergence** To check whether the convergence of $$f_n(x)$$ to $$f(x)$$ is uniform, recall the definition: convergence is uniform if, for any $$\epsilon > 0$$, there exists an $$N \in \mathbb{N}$$ such that for all $$n \geq N$$ and for all $$x \in [0, 1]$$, $$|f_n(x) - f(x)| < \epsilon.$$ 1. Fix $$\epsilon > 0$$. Consider $$|f_n(x) - f(x)|$$: - For $$x = 1$$, $$f_n(x) = 1 = f(x)$$, so the difference is $$0$$, which is less than $$\epsilon$$. - For $$0 \leq x < 1$$, $$f(x) = 0$$, so: $$|f_n(x) - f(x)| = |x^n - 0| = x^n.$$ The maximum value of $$x^n$$ for $$0 \leq x < 1$$ occurs at $$x = x_\text{max} = 1 - \frac{1}{n}$$ (where $$x^n$$ decreases fastest near 1). For large $$n$$, this value $$x^n_\text{max} \approx (1 - \frac{1}{n})^n \to e^{-1} \neq 0$$. 2. This implies that there exists no $$N$$ such that $$x^n < \epsilon$$ uniformly for all $$x \in [0, 1]$$, because $$x^n$$ does not become uniformly small on $$[0, 1]$$. #### **Conclusion (Uniform Convergence):** The convergence of $$f_n(x) = x^n$$ to the limit function $$f(x)$$ is **not uniform** on $$[0, 1]$$. --- ### **Summary:** 1. The pointwise limit of $$f_n(x) = x^n$$ on $$[0, 1]$$ is: $$f(x) = \begin{cases} 1 & \text{if } x = 1, \\ 0 & \text{if } 0 \leq x < 1. \end{cases}$$ 2. The convergence is **not uniform** on $$[0, 1]$$, because $$x^n$$ does not uniformly approach $$f(x)$$ as $$n \to \infty$$. --- Would you like a graphical demonstration of the convergence, or further details about uniform convergence? Here are 5 related questions to deepen your understanding: 1. How does the concept of uniform convergence differ from pointwise convergence mathematically? 2. What are other examples of sequences of functions that fail to converge uniformly on their domain? 3. How does the supremum norm play a role in checking uniform convergence? 4. Could the convergence of $$x^n$$ be uniform if the interval were $$[a, 1]$$ where $$a > 0$$? 5. How would the result change if $$f_n(x)$$ were defined on a semi-infinite domain, such as $$[0, \infty)$$? **Tip:** Uniform convergence ensures the interchangeability of limits and integrals/summations, a property not guaranteed with only pointwise convergence!