Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
szereg funkcyjny, I lilke it! –  AB_ 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

3 Answers 3

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

share|improve this answer
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

share|improve this answer
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 –  AB_ Mar 26 '13 at 8:22
@Q.matin and this function which is its convergence doesn't necessary have to be continuous (smooth) –  AB_ Mar 26 '13 at 8:26
@Q.matin you are very welcome –  AB_ 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 $1/n<x$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.