# Zero to the zero power - Is 0^0=1?

Dear All,

We have been having very good time in Algebra 1 classes. It is going very smooth and fun. In one of the lesson, we have discussed "What is the zero to the zero power? Is it 1,0, or undefined?" Some students brought impressive proof.

Here are some proof:

Clever student:

I know!

$x^{0}$ =  $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$ = $0^{1+x-1}$ = $0^{1} \times 0^{x-1}$ = $0 \times 0^{x-1}$$0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$.

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

$\lim_{x \to 0^{+}} \exp(x \log(x))$

$\exp( \lim_{x \to 0^{+} } x \log(x) )$

$\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

$\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

$\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

$\exp( \lim_{x \to 0^{+} } -x )$

$\exp( 0)$

$1$

So, since  $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

Best Regards,

Mustafa Yigit

Math Teacher and Math Olympiad Coach