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For any b in R, prove lim(b/n) = 0

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    For any b in ℝ, prove lim(b/n)=0


    2. Relevant equations
    I'm going to use the limit of a sequence involving any ε>0 for this.


    3. The attempt at a solution
    Let b be in ℝ
    Let ε>0, the 1/ε>0
    Let K be in N s.t. K>1/ε
    If n≥K, n≥1/ε
    This implies |b/n - 0| = |b|*|1/n - 0| = |b|*|1/n| < |b|*ε. Since ε>0 is arbitray, we can conclude the lim(b/n) = 0 for any b in ℝ.
     
  2. jcsd
  3. Mar 12, 2012 #2
    I'm not so sure you picked your big K correctly.

    I think what you are trying to prove is

    [itex]\displaystyle\lim_{n \rightarrow \infty} \frac{b}{n} = 0[/itex]

    So, by definition,

    [itex]\forall_{\epsilon > 0} \exists_{N \in \mathbb{N}}[/itex] such that for [itex]n \geq N, \quad |\frac{b}{n} - 0| < \epsilon[/itex]

    Solve for n in [itex]|\frac{b}{n} - 0| < \epsilon[/itex] and retry your proof!
     
  4. Mar 12, 2012 #3
    Thank you for your help. I was wrongly thinking I had to use cases due to b being negative. But how about this:

    Here is some scratch first:

    |b/n - 0| = |b/n| < ε So choose K>|b/ε|.

    Let ε>0.
    For all n≥K, |b/n - 0|=|b/n|<|b/(b/ε)|<ε.
    Since ε>0 is arbitrary, lim(b/n) = 0 for all b in R.

    How is that?
     
  5. Mar 12, 2012 #4
    You're almost there! A few things though:

    [itex]|\frac{b}{n}| < ε[/itex] So choose [itex]K>|\frac{b}{ε}|.[/itex]

    You should simplify your use of absolute values as much as possible so it doesn't make a mess. We know [itex]n > 0[/itex] since [itex]n \geq K \in \mathbb{N}[/itex]. So you can drop that absolute value sign and solve for n. Likewise, we assume [itex]\epsilon > 0[/itex], so there is no need for epsilon to be in absolute value.

    The only variable, in this problem, that needs to be in absolute value is b. Can you try modifying it now? It's almost correct, but you skip a bunch of steps and I'm not sure if you know exactly what you are solving and substituting.
     
  6. Mar 12, 2012 #5
    I'm basing this method a bit off of what my teacher used to write a "nice flowing proof" for more specific proofs. I'm trying to find the extra fat to cut off. I can understand that for every ε>0 we want to show the absolute value of |b/n - 0| < ε for all n ≥ K

    Sorry about skipping steps. I'll show you the scratch work that I left out.

    Scratch work is this part:

    |b/n - 0| = |b/n| = |b|/n < ε (I didn't do this third equality on the last problem because I just didn't really think about being explicit with n and ε > 0 . So if we solve for it, we will choose our K = |b|/ε. Thus for every n≥K, |b/n - 0| < ε.

    So for my proof I was using our class example and a substitution as:

    "The nice flowing proof"
    Let ε>0, choose K in N s.t. K>|b|/ε
    Then for all n≥K, |b/n - 0| = |b|/n ≤|b|/K < |b|/(|b|/ε) = ε
    Thus, lim(b/n) = 0 for all b in R. q.e.d

    This is from the substitution of |b|/ε for K.

    I see how this cleans it up a bit with the absolute value symbols!

    I'm sorry about not using the n-> infinity. The lim(b/n) is the convention my book chooses to use.
     
  7. Mar 12, 2012 #6
    I like it, the proof looks good! It seems picky, but if you better at writing neater proofs, it will help you in the long run when you don't have nice simple equations like this. Someone else will have to verify (I'm just an undergrad math major) but I see nothing wrong with that proof, nice work!

    Only thing I might change is not setting [itex]K = \frac{|b|}{\epsilon}[/itex] but instead letting [itex]K > \frac{|b|}{\epsilon}[/itex]. This is because you let [itex]n \geq K[/itex] which allows for the possibility of [itex]n = K[/itex].

    Everything else looks perfect though!
     
  8. Mar 12, 2012 #7
    Thank you very much!

    I believe this is correct :) I'm in undergrad analysis. Have you finished that already?

    I meant to write > as well--just didn't when typing it up. I would have written that part correctly on paper.
     
  9. Mar 12, 2012 #8
    Yeah, I finished that course 2 or years ago. I'm just finishing up this last semester and I have my B.S. Unfortunately that was one of my first proof classes and I didn't do as well as I had hoped to do. I was better when Analysis II rolled around.
     
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