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For each natural number $n \ge 2$, define:

$$f_n(x) = \frac{1}{1+x^n}$$

Find the function $f: [2, \infty) \to \Bbb R$ to which the sequence $\{f_n: [2, \infty) \to \Bbb R\}$ converges pointwise.

Answer: For any $x \ge 2$, $\lim_{n\to\infty} 1/(1 + x^n) = 0$, because $\lim_{n\to\infty} (1 + x^n) = \infty$.

Hence, $\{f_n (x)\}$ converges to $f(x) = 0$ pointwise for all $x \in [2, \infty)$.

how would we show if the convergence is uniform or not?

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$$ \forall x \ge 2: \left| \frac{1}{1+x^n} \right| \le \frac{1}{1+2^n} $$

The RHS goes to $0$ as $n \to \infty$. Hence, for any $\epsilon > 0$, we can find $N \in \Bbb N$ so that $n > N \implies \left|f_n(x)\right| < \epsilon \; \forall x \ge 2$. Uniform convergence follows.

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