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The sequence of continuous functions$$f_n(x) = \begin{cases} -1 & x\in [0, 1/2 - 1/n] \\[6pt] n(x - 1/2) & x\in [1/2 - 1/n, 1/2 +1/n] \\[6pt] 1 & x\in [1/2 +1/n, 1] \end{cases}$$

Converges to the function. $$f(x) = \begin{cases} -1 & x\in [0, 1/2 ) \\[6pt] 0 & x = 1/2 \\[6pt] 1 & x\in (1/2 , 1] \end{cases} $$

My question is that How can we get $f$? How this sequence of functions converges to $f$? How it comes $0$. Please explain me, I am not getting this.

Michael Hardy
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aryan
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  • Please define $f_n(x)$ explicitly. – Eclipse Sun Sep 19 '15 at 11:58
  • So, the limit of a function is usually a function, but in this case, the only domain for the limit function is the single point $\frac12$. Is this the problem exactly as it was presented to you? – Thomas Andrews Sep 19 '15 at 11:59
  • @ThomasAndrews Sir Exact Problem is this http://math.stackexchange.com/questions/152233/showing-c0-1-d-1-is-not-a-complete-metric-space – aryan Sep 19 '15 at 12:02
  • If the $f_n$ are the same as in that problem, the the limit is not $0$, except at $x=1/2$, where it is obviously zero. – Thomas Andrews Sep 19 '15 at 12:09
  • @ThomasAndrews Sir would you like to explain that How we calculate that $f$? I am totally confused. – aryan Sep 19 '15 at 12:11
  • Since you haven't asked a clear question, I have no doubt you are confused. You asked why it is zero, and I note it is not zero. You ask about $f$, but there is no $f$ in your question. What are you really trying to understand? – Thomas Andrews Sep 19 '15 at 12:14
  • @ThomasAndrews wait a minute Sir I am editing the question. – aryan Sep 19 '15 at 12:17
  • @ThomasAndrews Now see. – aryan Sep 19 '15 at 12:30
  • People do some incredibly weird things with MathJax code on this site, but this is the first time I've seen \text{} with nothing inside the \text{} except math notation enclosed in dollar signs. You could have just written the mathematical notation. \text{} has a purpose, and that is certainly not its purpose. I changed it. (Also, I replaced "array" with "cases".) ${}\qquad{}$ – Michael Hardy Sep 19 '15 at 18:21

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This is how you get f (you go point by point). you watch both how is $f_n$ defined and where, when $n\rightarrow\infty$ interval $[0,1/2-1/n]\rightarrow [0,1/2)$ , so your limit function f is equal to -1 on $[0,1/2)$ . now for the second case $[1/2-1/n,1/2+1/n]\rightarrow \{1/2\}$ when $n\rightarrow\infty$,this is single point and $f(1/2)=n\cdot 0 =0$ now for the last case $[1/2+1/n,1]\rightarrow (1/2,1]$ and $f=1$, that all .

jack
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