1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please correct my mistake in this limits exercise

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data
    we have this function

    f(x)=1 if [itex]\frac{1}{x}[/itex][itex]\in[/itex] 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. jcsd
  3. Feb 17, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Feb 17, 2013 #3
    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
     
  5. Feb 17, 2013 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  6. Feb 17, 2013 #5
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook