Proof that N to the power 0 is 1 using square roots

Question

I wondered if using square roots to prove that $N^0=1$ is valid (where $N$ is any real number). The way I propose to do this is as follows:

We consider when $x > 0$. If we do an iterative method:

$x_2=\sqrt{x_1}$

$x_3=\sqrt{x_2}$

and so on... we get that this would tend towards 1, so $x_n=1$ as $n \rightarrow \infty $.

This is equivalent to writing $$(x)^{\frac{1}{2}*\frac{1}{2}*...*\frac{1}{2}}$$ which tends towards $(x)^0$, and we know that this tends towards 1.

Would this be valid? And how would one prove this for $x \leq 0$, and perhaps formulate it better than I have managed to?

Answer 1

Assuming that you are using one of the inequalities

$$n\le\sqrt n\le1\text{ or }1\le\sqrt n\le n$$

to squeeze, you are indeed showing that

$$\lim_{x\to0}n^x=1,$$ if the limit exists.

But

• this is not sufficient to prove that the limit exists (as you just use the particular exponents $x=2^{-k}$),

• this does not "prove" $n^0=1$, which is a pure matter of convention, but proves that the function $n^x$ is continuous at $0$ when you admit that $n^0:=1$.