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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?

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    $\begingroup$ $\infty$ is not in $(0,\infty)$. Thus your question is irrelevant. $\endgroup$ – Chris Eagle Jul 11 '13 at 13:35
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$0^t=0 \lt 1 = t^0$ for all $t \in (0,\infty)$.

So the answer would be correct if it showed $t^0$ was positive.

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  • $\begingroup$ $ o^{0}=1 $ since $ x^{x}=e^{xln(x)} $ and $ x \to 0 $ $ xln(x)=1 $ $\endgroup$ – Jose Garcia Jul 11 '13 at 13:50
  • $\begingroup$ @JoseGarcia: $0 \not \in (0, \infty)$ $\endgroup$ – Henry Jul 11 '13 at 14:05
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    $\begingroup$ @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 $x=0$. $\endgroup$ – celtschk Jul 11 '13 at 14:19

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