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Let $h_n$ be defined on the interval $\mathbb{I}=[0,1]$ by the formula $$h_n(x)=\begin{cases} nx, & 0\le x\le 1/n \\\\ \frac{n}{n-1}(1-x), &1/n<x\le1. \end{cases}$$

Show, by defintion, that $\lim(h_n)$ exists on $\mathbb{I}$.

I know just by looking at it if $x=0$ it converges, but I do not know how to prove this using the defintion? What is giving me problems is the way it is written. I haven't dealt with a limit proof question written like this before.

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szereg funkcyjny, I lilke it! – where_is_tftp Mar 26 '13 at 7:52
I think it converges to the function $1-x$ if $x\neq 0$. Choose $\epsilon >0$, then let $n$ be big enough so that $1/n<x\leq 1$, so you know you will use $\frac{n}{n-1}(1-x)$ and you can make this arbitrarly close to $1-x$ because $|(1-x)-\frac{n}{n-1}(1-x)|=\frac{1}{n-1}(1-x)$, and for a fixed $x$, the quantity $1-x$ is constant. – Kanye West Mar 26 '13 at 7:53
@cf16 Huh? You must be speaking polish. – Q.matin Mar 26 '13 at 7:55
@KanyeWest I never knew Kanye West was a math guy. Thanks for that hint! – Q.matin Mar 26 '13 at 7:57
@Q.matin Im a musical genius! The voice of a generation! – Kanye West Mar 26 '13 at 7:58
up vote 1 down vote accepted

For all $x>0$, if $n\gt\frac1x$, then $h_n(x)=\frac{n}{n-1}(1-x)$. Therefore, if $x\gt0$, then $\lim\limits_{n\to\infty}h_n(x)=1-x$.

For all $n$, $h_n(0)=0$. Thus, $$ \lim\limits_{n\to\infty}h_n(x)=\left\{\begin{array}{} 1-x&\text{if }x\gt0\\ 0&\text{if }x=0 \end{array}\right. $$

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Thanks Rob, this is much more clear. Last question, Kayne in the comments related to $|(1-x)-\frac{n}{n-1}(1-x)|=\frac{1}{n-1}(1-x)$ by the defintion. How would I complete that? – Q.matin Mar 26 '13 at 8:52
Choose an $\epsilon\gt0$. Let $n\gt\max\left(\frac1x,1+\frac1\epsilon\right)$, then $\left|h_n(x)-(1-x)\right|=\frac1{n-1}(1-x)\le\epsilon$. – robjohn Mar 26 '13 at 9:11
Thanks a lot !! – Q.matin Mar 26 '13 at 20:18

for every x in [0,1] there exists n that $x> \frac{1}{n}$ so $\lim(h_n)(x)=\lim_n_{\infty}\frac{n}{n-1}(1-x)=1-x$ since x is constant

for x=1 and x=0 lim=0

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except at $x=0$ where $\lim\limits_{n\to\infty}h_n(0)=0$ – robjohn Mar 26 '13 at 8:15
Thanks, I see why it converges to $1-x$ but how can I prove this using the defintion or will your answer be suffice? – Q.matin Mar 26 '13 at 8:17
@Q.matin yes, definition states that this series converges pointwisely to function $f_n$ if it converges in every point – where_is_tftp Mar 26 '13 at 8:22
@Q.matin and this function which is its convergence doesn't necessary have to be continuous (smooth) – where_is_tftp Mar 26 '13 at 8:26
@Q.matin you are very welcome – where_is_tftp Mar 26 '13 at 9:03

Hint: Since you already know what's going on at $x=0$, fix some non-zero element of the interval and then choose $n\in\mathbb N$ large enough so that $1/n<x$.

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What $n$ should I choose? – Q.matin Mar 26 '13 at 8:05
@Q.matin Any $n$ that satisfies the condition. – Mark McClure Mar 26 '13 at 8:06
I am not following, can you elaborate further please? – Q.matin Mar 26 '13 at 8:11

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