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Prove that the limit constant times a function

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    the question is: Prove that if lim[itex]_{x→0}[/itex] f(x)/x = l and b≠0, then lim[itex]_{x→0}[/itex] f(bx)/x = bl

    2. Relevant equations
    Hint: write f(bx)/x = b[f(bx)/bx]
    properties of limits, delta epsiolon...

    3. The attempt at a solution
    I assume that I can use the property of limits that is lim(x->a) (f*g)(x) = l*m
    I can make b one function and f(x)/x another, and hence use the above property. But I honestly don't think that's right, don't know where to go from here
     
  2. jcsd
  3. Oct 10, 2012 #2

    SammyS

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    Where to go from here? ... good question.

    Do you suppose you are to do an ε - δ proof, or can you do something with substitution?
     
  4. Oct 10, 2012 #3
    I think that would suffice, after all x is an infinitesimal and multiplied by a constant won't change anything. Just like infinitesimal is a number smaller than any given number ε, so it is perfectly okey to have an infinitesimal smaller than [itex]\frac{ε}{b}[/itex] I think...
     
    Last edited: Oct 10, 2012
  5. Oct 10, 2012 #4
    You could use ε - δ proof, but can also use the properties of limits to prove this, like sum of functions is equal to sum of their limits, and product and quotient properties, at least that's what my book says. I don't get the hint given in my book though.

    can you explain that further, I'm kind of new to all this, only in first year.
     
  6. Oct 10, 2012 #5

    HallsofIvy

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    Let u= bx so that x= u/b. Then
    [tex]\frac{f(bx)}{x}= \frac{f(u)}{\frac{u}{b}}= \frac{bf(u)}{u}= b\frac{f(u)}{u}[/tex]
    and, of course, as x goes to 0, so does u.
     
  7. Oct 10, 2012 #6
    I don't get it :cry: also what happens if b=0
     
  8. Oct 10, 2012 #7
    If I remember it correctly, an infinitesimal is a quantity that is smaller than any number given, hence here [itex]lim_{x→0} x[/itex] is smaller than an arbitrary |ε|. where ε can be as small as 0.000000000000001 or -0.000000000000000000000001. So as we can see when x is indefinitely small, if multiplied by a constant b, it is still very very small(the quantity won't change much if you multiply 0.0000000000000000001 by 6 for example), so we can redefine [itex]lim_{x→0} x[/itex] as smaller than [itex]|\frac{ε}{b}|[/itex](ε is still any arbitrary value and clearly it is equivalent with the above definition). Or in a less dogmatic sense, we can say [itex]lim_{x→0} x[/itex] is very small(can be regarded as 0) so multiply it with any constant it is still almost zero(0 multiplies with any number still gives 0), and then we can use the substitution to obtain the results as HallsofIvy said.

    If b = 0, then for any values of x, f(bx) degrades into f(0).
     
  9. Oct 10, 2012 #8
    for example, [itex]lim_{x→0}\frac{sin bx}{x} = b[/itex] for b is not zero. If b is 0, then the whole expression becomes [itex]lim_{x→0}\frac{sin 0}{x} = 0[/itex](because anything multiplies zero only gives zero). You can do a Taylor expansion to verify this I think.
     
  10. Oct 10, 2012 #9
    Okay now I get it, after contemplating for a long time, lol thanks.

    Yes and thanks for the help!
     
  11. Oct 10, 2012 #10

    Mark44

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    In your first post, you say that b ≠ 0.

    Famous saying:
     
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