# Please correct my mistake in this limits exercise

1. Feb 17, 2013

### Andrax

1. The problem statement, all variables and given/known data
we have this function

f(x)=1 if $\frac{1}{x}$$\in$ Z ( aka integer)
f(x) = 0 otherwise
Prove that limit (as x approach 0) dosen't exist (use the definition of limit - trying to prove that limit as x approchaes 0 and x approchs 0+ will not work here)<-- hint given by the exercise

2. Relevant equations

3. The attempt at a solution
so what i've done here , first of all this is obviously done by contradiction but just please correctr my mistaker here
let x=1/p /p is an integer (aka 1/x is an integer) then f(x) = 1
we have limit (as x approachs 0 ) 1 = 1
now let x=/ 1/p /p s an integer then
f(x)=0
then limit as ( x approachs 0 ) 0 = 0
we haven limit as ( x approachs 0 ) 0 and 1 which is impossible then it's a contradition please correct me here i know my proof is wrong but i couldn't correct it

2. Feb 17, 2013

### jbunniii

The idea behind your proof is basically OK but you need to state it more clearly.

If $\lim_{x \rightarrow 0}f(x)$ exists, then we must have $\lim_{n \rightarrow \infty}f(x_n) = \lim_{n \rightarrow \infty}f(y_n)$ for any sequences $(x_n)$ and $(y_n)$ such that $x_n \rightarrow 0$ and $y_n \rightarrow 0$.

So one way to prove that $\lim_{x \rightarrow 0}f(x)$ does not exist is to find two sequences $(x_n)$ and $(y_n)$ such that $x_n \rightarrow 0$ and $y_n \rightarrow 0$ but $\lim_{n \rightarrow \infty}f(x_n) \neq \lim_{n \rightarrow \infty}f(y_n)$.

As I think your proof is indicating, we can choose $x_n = 1/n$, in which case $f(x_n) = 1$ for all $n$, so $\lim_{n \rightarrow \infty}f(x_n) = 1$.

Now construct a sequence $(y_n)$ that will give you the contradiction.

3. Feb 17, 2013

### Andrax

thank you this is a way to prove it but i'll stick for a couple of minutes with the definition of limits i think if f(x) = 1 or 0 then limit (x->0) must be 0 or 1
if i prove that it isn't 0 and it isn't 1
i'll try to set L = 0 in the first case and look for a contradiction

4. Feb 17, 2013

### jbunniii

With this, you can rule out L = 0 or L = 1, but unless you prove that these are the only possibilities, that doesn't finish the job. However, if you assume that the limit exists and equals some general L, taking $\epsilon < 1/2$ should allow you to reach a contradiction.

5. Feb 17, 2013

### Andrax

i've managed to get a contradiction by the definition i did the opposite of it and i gave epsilon , and alpha which made the contradition for both l= 0 and L = 1 ( i got it by f(x) 0> 1/2 )
thank you