# Find the limit function [on hold]

Suppose that for $n \in \mathbb{N}$, $$f_n(x) = \begin{cases} -1, & -1 \leq x \leq - 1/n, \\ nx, & |x| \leq 1/n, \\ 1, & 1/n \leq x \leq 1. \end{cases}$$

a) Find the limit function for $\{f_n\}$.

b) Determine whether $\{f_n\}$ converges uniformly on $[-1,1]$.

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## put on hold as off-topic by Jonas Meyer, Tunk-Fey, 900 sit-ups a day, Jyrki Lahtonen, J. W. PerryAug 18 at 5:59

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Tunk-Fey, 900 sit-ups a day, Jyrki Lahtonen, J. W. Perry
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\me speechless –  Gigili May 29 '12 at 18:46
Try drawing the function for a few small values of $n$. And your math formatting needs just a little bit of work, hence all the downvotes, I assume. –  Dustan Levenstein May 29 '12 at 18:50
Please...do edit ASAP your question, if possible using LaTeX, but at least paying attention to the parentheses and spacing between words... –  DonAntonio May 29 '12 at 18:52
@DustanLevenstein The downvotes are for the constant avoidance of our comments trying to help him format his questions in terms of wording. It is always in a command-fashion - telling us to do something without any help on his part. How did he encounter the problem, what does he know so far, etc.? For examples, see ALL of his past questions. –  Joe May 29 '12 at 18:59
@Thomas: Well, I've tried enough times (not me specifically, the community) to guide this user. since I see no changes, I wouldn't bother being supportive. –  Gigili May 29 '12 at 19:16

Have you observed that for almost all $\,x\in [-1,1]\,$ there exists a natural index from which on the function's value is $\,1\,$ ...what's this value that no doubt spoils the uniform convergence?
a) $f_n(x) \rightarrow f(x)$, a.e where $$F(x) = \left\{ \begin{array}{rcl} -1,& \mbox{se} & x < 0\\ 0, & \mbox{se} & x=0\\ 1, & \mbox{se} & x > 0 \end{array} \right.$$ In fact, $f_n(0) = 0 \ \forall n$ and if $x \neq 0$. Chosse $n_0$ such that $1/n_0 < |x|$. Then $$n \ge n_0 \Rightarrow |f_n(x) -1| = 0 < \epsilon$$ for any $\epsilon$. b) If $f_n$ converges uniformily, by a) it converges to $f$ above. But this is not true because for $\epsilon = 1/4, \forall n$ take $x = 1/2n$. Then $$|f_n(x) - 1| = |n 1/2n - 1| = 1/2 > 1/4.$$