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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm stuck on a homework problem

For each natural number $n$ and each number $x \in [0,1]$ let $$f_n(x) = \frac{x}{nx+1}$$ Find the function $f:[0,1] \to \mathbb{R}$ to which the sequence $\{f_n : [0,1] \to \mathbb{R}\}$ converges pointwise. Prove the convergence is uniform.

Any ideas? Thanks!

share|cite|improve this question
Where did you get stuck? Do you know what pointwise convergence means? Did you try to figure out the limit of $f_n(1)$, say? Do you know what uniform convergence is? – Dirk Dec 2 '11 at 8:09
up vote 1 down vote accepted

$f_n$ clearly converges to $g(x)=0$ pointwise.

Indeed, $f_n(0)=0$ for all $n$. While, for $0<x\le1$, we have ${1\over x}\ge1 $; whence:

$$ \tag{1}|f_n(x)|=| {x\over nx+1}| ={1\over n+{1\over x}}\le{1\over n+1}\ \buildrel {n\rightarrow\infty}\over{\longrightarrow} \ 0. $$

The above actually shows that the convergence is uniform: we can make $f_n(x)$ small for all $x$ by taking $n$ sufficiently large.

To be formal:

Let $\epsilon>0$. Choose $N$ so that ${1\over N+1}<\epsilon$. Then if $n\ge N$, we have, using (1): $$|f_n(x)-0|=|f_n(x)|\le{1\over n+1} \le{1\over N+1}<\epsilon$$ for all $0< x\le1$. Also, $|f_n(0)|=0<\epsilon$. Thus, the convergence is uniform.

share|cite|improve this answer
Hmm, isn't there a policy for questions about homework here… ? – Dirk Dec 2 '11 at 9:55

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.