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It seems that $f_n(x)=1-x^n$ converges to $f(x)$ such that \begin{eqnarray*} f(x)&=&1~~\text{for}~~x\in[0,1)\\ &=& [0,1]~~\text{for}~~x=1.\end{eqnarray*}

I wonder if it is correct or not. If it is, how to show this convergence?

user722137
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  • $x^n$ converges to $0$ for $|x|<1$ and to $1$ for $x=1$. In the other cases, it diverges. – bjorn93 Dec 23 '19 at 14:37
  • How are you getting this limit at 1? – cmk Dec 23 '19 at 14:38
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    Surely $f(1)$ is identically equal to $0$ for all $n$? How on earth would you say it is $[0,1]$? What would that even mean? – MPW Dec 23 '19 at 14:39
  • Thanks for the comments and answers. What I was thinking is not correct in the end. The reason why I was thinking in this way is the graph of $f_n(x)$ with a large $n$ looks like $f(x)$ defined above. – user722137 Dec 24 '19 at 10:28

1 Answers1

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Note that for $0\le x<1$ we have $\lim _{n\to \infty} x^n =0$ therefore $1-x^n \to 1$

On the other hand for $x=1$ we have $1-x^n=0$ so the limit is $0$

Mohammad Riazi-Kermani
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