91Ó°ÊÓ

To find the pointwise limit of the sequence \(f_{n}\) on \([0,1]\), we substitute each \(x\) value in the range into the function and find the limit as \(n\) approaches infinity: \(\lim_{n \to \infty} \frac{1}{(1+x)^n}\). For \(x \in [0,1)\), the limit is 0 while for \(x=1\), the limit is 1.

To check for uniform convergence, we examine whether for each \(\epsilon > 0\), there exists \(N \in \mathbb{N}\) such that for all \(n \geq N\) and all \(x\) values in the range, the inequality \(|f_{n}(x)-f(x)| < \epsilon\) holds. Given that \(f_{n}(x)\) converges to 0 for all \(x \in [0,1)\) and to 1 when \(x=1\), we have \(f_{n}(x)\) converging to \(f(x)\) with |\(f_{n}(x) - f\) | = |\(f_{n} - 0\) | = \(f_{n}\) for \(x \in [0,1)\) and |\(f_{n}(x) - f\) | = |\(f_{n} - 1\) | = | \(1/(1+1)^n - 1 \)| for \(x = 1\). We can see that there does not exist \(N \in \mathbb{N}\) such that for all \(n \geq N\), \(|f_{n}(x) - f(x)| < \epsilon\) for all \(x\) in the range as we have different behavior at \(x=1\). So, the sequence does not converge uniformly to \(f\) on \([0,1]\).