Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits of frac(x)/x

  1. Sep 4, 2015 #1
    $$\lim_{x\to\infty} \frac {frac(x)} {x} $$
    frac(x) is x minus floor function of x. So if x = 2.5, floor function = 2 and frac(x) = 0.5
    Hence frac(x) will always be a number between -1 and 1 but never -1 and 1.

    By squeeze theorem,
    -1 < frac(x) < 1
    -1/x < frac(x)/x < 1/x
    0 < frac(x)/x < 0

    Does this means that $$\lim_{x\to\infty} \frac {frac(x)} {x} = Undefined? $$
    Since it is between 0 but not 0.
    However WolframAlpha gives the answer as 0.
    Shouldn't it be 0 only if it is $$0 \leqslant frac(x)/x \leqslant 0$$
    So did I do something wrong or is WolframAlpha wrong?
     
  2. jcsd
  3. Sep 4, 2015 #2

    Titan97

    User Avatar
    Gold Member

    No. frac{x} is always positive and lies between 0 and 1.

    In the limit, the numerator always lies between 0 and 1 while the denominator is an increasing function that tends to infinity when x tends to infinity.

    So what will be ##\lim_{x\to\infty}\frac{frac(x)}{x}##
     
  4. Sep 5, 2015 #3
    Thank you Titan97 for clearing that up. So the limits would be 0 this time.
     
  5. Sep 5, 2015 #4

    Mark44

    Staff: Mentor

    You can use the squeeze theorem here since ##0 \le \text{frac(x)} < 1##.
     
  6. Sep 5, 2015 #5

    Titan97

    User Avatar
    Gold Member

    Yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limits of frac(x)/x
  1. Sinx /x limit (Replies: 1)

Loading...