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let $$f_n(x)=\begin{cases} 1-nx&\text{when }x\in[0,1/n]\\0&\text{when }x\in [1/n,1]\end{cases}$$ Which of the following is correct?

  1. $\lim_{ n\to\infty} f_n(x)$ defines a continuous function on $[0,1]$
  2. $\{f_n\}$ converges uniformly on $[0,1]$
  3. $\lim_{n\to\infty} f_n(x)=0$ for all $x\in [0,1]$
  4. $\lim_{n\to\infty} f_n(x)$ exists for all $x\in[0,1]$
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Welcome to MSE! It really helps readability to format questions using MahtJax (see FAQ). regards – Amzoti Jul 23 '13 at 17:19
fn(0)=1 and fn(1)=0, so definitely it's not uniform convergence and neither lim n->infinity fn(x)=0 for all x. so, we are left with 'a' and 'd' options. – Ramit Jul 23 '13 at 17:21
@Ramit Yes. Are you sure that $1/n$ is included on both parts of the piecewise definition? If yes, continuity is apparent. – Torsten Hĕrculĕ Cärlemän Jul 23 '13 at 17:33
@TorstenHĕrculĕCärlemän yes, 1/n is there in both parts. closed bracket, not open. but the answer is 'd' as per the key that I am checking. – Ramit Jul 23 '13 at 17:48
@Hagen- thanks for editing my question. – Ramit Jul 23 '13 at 17:50
up vote 3 down vote accepted

Let denote $\displaystyle f=\lim_{n\to\infty} f_n$.

For $x=0$ we have $f_n(0)=1,\quad\forall n>0$ so $f(0)=1$.

For $x>0$ there's $N\in\mathbb N$ such that $\frac{1}{n}\leq x,\quad \forall n\geq N$ so $f_n(x)=0\quad \forall n\geq N$ and then $f(x)=0$ so we conclude: $$f(x)=\left\{\begin{array}\\ 1&\text{if}\ x=0\\ 0&\text{if}\ 0<x\leq1 \end{array}\right.$$

Now can you answer the questions?

It's clear that $f$ isn't continuous at $0$ so options 1. and 2. are false and $f(0)=1$ so also option 3. isn't true and since $f(x)$ exists for all $x\in[0,1]$ so option 4. is true.

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thanks @Sami for this part. But I am yet not able to tick the correct option. I am sensing that the limit doesn't exist on 0, since function value and right limit don't match. so it's not continuous. very obviously, m thinking wrong. looks like m not clear on continuity. i'll try to refer some book. but i'll appreciate if u can provide some quick help. – Ramit Jul 23 '13 at 17:55
I edited my answer. – user63181 Jul 23 '13 at 18:03
Limit at $0$ of what? – user63181 Jul 23 '13 at 18:12
lim n->infinity f(0)=1 lim n->infinity f(0.5)=0 lim n->infinity f(1)=0 clearly, limit exits for all. but why the discontinuity? I wish u could explain. thanks. – Ramit Jul 23 '13 at 18:15
You can see that $f$ isn't continuous at $0$ since $\lim_{x\to 0^+} f(x)=1\neq f(0)=0$ and since the functions $f_n$ are continuous and their pointwise limit $f$ isn't continuous so the convergence isn't uniform. – user63181 Jul 23 '13 at 18:20

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