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Homework Help: Prove Limit of x^3sin(1/x)

  1. Feb 5, 2014 #1
    1. The problem statement, all variables and given/known data


    Limit of x3sin(1/x)=0 as x→0

    3. The attempt at a solution

    Intuitively, I understand that since sin(1/x) always lies between 1 and -1, and since x approaches zero, x3sin(1/x) must also approach zero. When formally expressing proof, however, which do we restrict, x3 or sin(1/x)?
  2. jcsd
  3. Feb 5, 2014 #2


    Staff: Mentor

    There are two things I don't understand.
    1. How formally do you need to prove this? I.e., do you need to do a δ-ε argument, or can you use a theorem?
    2. What do you mean "which do we restrict"? The variable x is approaching zero, so both x3 and sin(1/x) are affected.
  4. Feb 5, 2014 #3
    Preferably delta-epsilon. I know how to prove it with squeeze theorem, but I got stuck at δ-ε proof.
    I think I have an idea of how to prove it with δ-ε, but I'm not sure.
    EDIT: Oh, and by restrictions, I mean what is taken as δ so we can focus on one function of x, since it's not multiplied by a constant, rather, another function of x.

    so if ε>0

    x3 |sin(1/x)|≤x3

    since |x3sin(1/x)| = |[x3sin(1/x)]-0| <ε and |x3| <ε

    let δ=cubic√ε
    |[x3sin(1/x)]-0| = |f(x)-0| <ε
  5. Feb 5, 2014 #4


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    Science Advisor
    Homework Helper
    Gold Member

    You definitely have the correct idea and a good argument. But you could write it up a bit nicer. Since you have already solved it, I will suggest how you should write up your argument:

    Suppose ##\epsilon > 0##. Pick ##\delta = \sqrt[3]\epsilon##. Then if ##0<|x-0|<\delta##$$
    \left| x^3\sin(\frac 1 x)\right| =\left| x^3\right|\left| \sin(\frac 1 x)\right| \le |x^3|\cdot 1
    < (\sqrt[3]\epsilon)^3=\epsilon $$Note that you can't allow ##x=0## in the argument.
  6. Feb 5, 2014 #5
    Awesome, yeah that's a lot cleaner and less redundant than mine. Thanks so much! Really appreciate it. :)
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