Proving Limit Definition: x^n = a^n for all positive integers n

[Icebreaker]

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

$$(x^{n/2}+a^{n/2})(x^{n/4}+a^{n/4})(x^{n/8}+a^{n/8})...(x+a)(x-a)$$

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?

 

[Oxymoron]

Couldn't you just write

$$\lim_{n\rightarrow a}x^n = \lim_{n\rightarrow a}(x\cdot x\cdot\dots\cdot x)$$

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

$$\lim_{n\rightarrow a}x \cdot \lim_{n\rightarrow a}x \cdot \dots \cdot \lim_{n\rightarrow a}x$$

again n times.

Then use the definition of a limit that

$$\lim_{n\rightarrow a}x = a$$

and conclude that

$$\lim_{n\rightarrow a}x^n = \lim_{n\rightarrow a}a\cdot \lim_{n\rightarrow a} \cdot \dots \cdot \lim_{n\rightarrow a}a$$

n times. Which is obviously

$$\lim_{n\rightarrow a}a^n$$

again by the algebra of limits.

 

[Icebreaker]

x->a, not n->a

 

[Hurkyl]

I could look for an upper bound, but it's unbounded.

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)

 

[Icebreaker]

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.

 

[Hurkyl]

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.

 

[Oxymoron]

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

$$\lim_{x\rightarrow a} x = a$$

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

$$\lim_{x\rightarrow a} x^n = a^n$$

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