zero raised to infinity

I encountered a question where the only condition stated that $t>0$ and was then asked to compare these two quantities

1. $0^t$
2. $t^0$

The scope of $t$ is $(0,\infty)$ and hence for infinity 1.) and 2.) are not defined . However the answer states that 2.) > 1.) as 1.) would always be zero . Is that the correct explanation?

• $\infty$ is not in $(0,\infty)$. Thus your question is irrelevant. – Chris Eagle Jul 11 '13 at 13:35

$0^t=0 \lt 1 = t^0$ for all $t \in (0,\infty)$.
So the answer would be correct if it showed $t^0$ was positive.
• $o^{0}=1$ since $x^{x}=e^{xln(x)}$ and $x \to 0$ $xln(x)=1$ – Jose Garcia Jul 11 '13 at 13:50
• @JoseGarcia: $0 \not \in (0, \infty)$ – Henry Jul 11 '13 at 14:05
• @JoseGarcia: With the same right I can say $0^0 = \lim_{x\to 0+0}0^x = \lim_{x\to 0+0}0 = 0$. You simply cannot unambiguously determine $0^0$ as limit because $(x,y)\mapsto x^y$ is not continuous at $(0,0)$. However for many problems it is meaningful to define $0^0=1$, for example so you can write polynomials as $\sum_{k=0}^na_kx^k$ without making a special case either for the term $a_0$ or for the value $x=0$. – celtschk Jul 11 '13 at 14:19