1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of Limit

  1. Oct 15, 2005 #1
    Prove using the def of lim that lim(x->a) x^n = a^n for all positive integers n.

    The limit is obviously a^n. I've factored x^n - a^n and got

    [tex] (x^{n/2}+a^{n/2})(x^{n/4}+a^{n/4})(x^{n/8}+a^{n/8})...(x+a)(x-a)[/tex]

    Now, I have my x-a, which is needed for determining delta, i.e., |x-a|<d. However, that long chain of factors before it I have no idea how to get rid of. I could look for an upper bound, but it's unbounded. Can anyone help?
  2. jcsd
  3. Oct 15, 2005 #2
    Couldn't you just write

    [tex]\lim_{n\rightarrow a}x^n = \lim_{n\rightarrow a}(x\cdot x\cdot\dots\cdot x)[/tex]

    Where there are n x's in that product. Then use the algebra of limits to split this into

    [tex]\lim_{n\rightarrow a}x \cdot \lim_{n\rightarrow a}x \cdot \dots \cdot \lim_{n\rightarrow a}x[/tex]

    again n times.

    Then use the definition of a limit that

    [tex]\lim_{n\rightarrow a}x = a[/tex]

    and conclude that

    [tex]\lim_{n\rightarrow a}x^n = \lim_{n\rightarrow a}a\cdot \lim_{n\rightarrow a} \cdot \dots \cdot \lim_{n\rightarrow a}a[/tex]

    n times. Which is obviously

    [tex]\lim_{n\rightarrow a}a^n[/tex]

    again by the algebra of limits.
  4. Oct 16, 2005 #3
    x->a, not n->a
  5. Oct 16, 2005 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No it's not.

    Incidentally, your approach only works when n is a power of 2. Try:

    p^n - q^n = (p-q) (p^{n-1} + p^{n-2}q + \cdots + p q^{n-2} + q^{n-1})

    (P.S. n->a was just a typo in Oxymoron's post. His approach works too)
    Last edited: Oct 16, 2005
  6. Oct 16, 2005 #5
    But using Oxymoron's approach, by the definition of a limit alone, can we say lim x^n = (lim x)^n? Or perhaps I misunderstood something.
  7. Oct 16, 2005 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, Oxymoron's approach uses the first properties of limits (in particular, that the limit of a product is the product of limits), so I guess it's not appropriate for your exercise.
  8. Oct 16, 2005 #7
    No. My approach uses both. You can use the algebra of limits to do the tedious work, but when it comes down to it, you still have to prove that

    [tex]\lim_{x\rightarrow a} x = a[/tex]

    then once you prove this from first prinicples (which is REALLY EASY) you know it works for

    [tex]\lim_{x\rightarrow a} x^n = a^n[/tex]

    which is just by the algebra of limits. So it uses both: properties of a limit AND the actual definition from first principles.

    PS. Sorry about the typo
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Definition of Limit
  1. Definition of Limit (Replies: 2)

  2. Definition of Limit (Replies: 3)

  3. Definition of Limit (Replies: 2)

  4. Definition of a limit (Replies: 3)

  5. Definition of a Limit (Replies: 1)