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Showing Continuity of a function

  1. Mar 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that f(x) = x/(1+x^2) is continuous on R


    2. Relevant equations
    f is continuous at a if for any epsilon > 0, there exists a number delta > 0 such that if |x-a|<delta, then |f(x)-f(a)|<epsilon.


    3. The attempt at a solution
    |f(x) - f(a)| = |x/(1+x^2) - a/(1+a^2)|

    = |(x-a)(1-ax)/[(1+x^2)(1+a^2)]|

    If I can show that the previous is <= |(x-a)(1-ax)/(1-ax)| then I can cancel the (1-ax) and I'll be done right? I'm not even really sure if I started on this the right way, so any help would be appreciated.
     
  2. jcsd
  3. Mar 23, 2008 #2
    one way to try to figure it out...

    Let e > 0. Choose d = ....don't know yet...
    Suppose |x-a| < d, then
    |x/(1 + x^2) - a/(1 + a^2)| = |(x -a + ax(a-x))/((1 + x^2)(1 + a^2))| <= |x-a|/|(1 + x^2)(1 + a^2)| + |ax||x-a|/|(1 + x^2)(1 + a^2)| <= |x-a|/(1 + a^2) + |ax||x-a|/(1 + a^2) <= |x-a| + |x/a||x-a| <= think about what you can say about |x/a|, then try to find a delta that works

    this leads to a proof, but there might be an easier way, play with it, goodluck
     
    Last edited: Mar 23, 2008
  4. Mar 23, 2008 #3
    ok I got it I think. If I set |x-a|<|a|, then |x|-|a|<|a| or |x/a|<2. So the original
    |f(x)-f(a)|<3|x-a|. So, if delta = epsilon/3 then the definition given in the first post will be satisfied.

    Side note, does that imply uniform continuity since delta depends only on epsilon?
     
  5. Mar 23, 2008 #4
    good work, but not quite

    first, if a = 0, everything is trivial(you should verify), so say a != 0

    so we have stuff <= |x-a| + |x/a||x-a|.

    as you pointed out, we need |x-a| < |a|. But we also need |x-a| < e/3, so choose
    d = min{|a|, e/3}, then if |x-a| < d (this means we have both |x-a| < d <= e/3 and |x-a|<d <= |a| )

    stuff <= |x-a| + |x/a||x-a| < e/3 + 2e/3 = e.
     
    Last edited: Mar 23, 2008
  6. Mar 24, 2008 #5
    Another method, which may be simpler:
    If a function f is differentiable at a point x then it must be continuous at x. Thus, it is enough to show that f is differentiable on the whole real line, i.e. that f '(x) exists for all real x.
     
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