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Homework Help: 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


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    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


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    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
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