25. Zero's zeroth power is 1 (An interesting observation)

We have: $$ \begin{align} 1^1&=1 &&& 0^1&=0 \\ 1^0&=1 &&& 0^0&=1 \end{align} $$ Are you surprised by $0^0=1$?

 

Proof. To prove this, we consider $$ \begin{align} y=x^x \end{align} $$ and take the limit $x\rightarrow0$. It is not straightforward to do this directly with $x^x$ so we take the (natural or any) logarithm on both sides:

$$ \begin{align} \ln y=\ln x^x=x\ln x \end{align} $$

Let $x=e^{-n}$ and the limit $x\rightarrow0$ is represented by $n\rightarrow\infty$, i.e.

$$ \begin{align} &&\lim_{x\rightarrow0}x\ln x &=\lim_{n\rightarrow\infty}e^{-n}\ln e^{-n} \\ &&&=\lim_{n\rightarrow\infty}(-n)e^{-n}\underbrace{\ln e}_{=1} \\ &&&=-\lim_{n\rightarrow\infty}\frac{n}{e^n} \\ &&&=0 \\ \\ \Rightarrow&& \lim_{x\rightarrow0}x^x&=\lim_{x\rightarrow0}e^{x\ln x}=e^0=1 \quad\checkmark \end{align} $$

 

Alternative. We may view $y=\ln x$ as:

$$ \begin{align} && \int\frac1x\,dx&=\ln\vert x\vert+c \\ \\ \Rightarrow&& \ln x&=\lim_{\alpha\rightarrow1}\int_1^x\frac1{t^\alpha}\,dt \\ &&&=\lim_{\alpha\rightarrow1}\frac1{1-\alpha}\left(x^{1-\alpha}-1\right) \\ \\ \Rightarrow&& \lim_{x\rightarrow0}x\ln x&=\lim_{x\rightarrow0}\left[\lim_{\alpha\rightarrow1}\frac1{1-\alpha}\left(x^{2-\alpha}-x\right)\right] \\ &&&=\lim_{\alpha\rightarrow1}\left[\lim_{x\rightarrow0}\frac1{1-\alpha}\left(x^{2-\alpha}-x\right)\right] \\ &&&=0 \quad\checkmark \end{align} $$

Exercise. For enthusiasts, uniform convergence is assumed for switching the two limits.