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diamong [38]
3 years ago
12

Define fn : [0,1] --> R by the

Mathematics
1 answer:
sasho [114]3 years ago
8 0

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

The limit \lim_{n\to \infty }c^{n} exists and converges to zero whenever \lvert c \rvert. But, if c=1 the sequence \{c^{n}\} is constant and all its terms are equal to 1, then converges to 1. Using this result, consider the sequence of functions \{f_{n}\} defined on the interval [0,1] by f_{n}(x)=x^{n}. Then, for all 0\leq x we have that \lim_{n\to \infty}x^{n}=0. Now, if x=1, then \lim_{n\to \infty }x^{n}=1. Therefore, the limit function of the sequence of functions is

f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

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