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I was reading this answer for the proof of $$\lim_{x\to a}{x^n}=a^n\ \forall \ n \in \mathbb{R}$$ Now..according to the answer given, the proof uses the following inequality $$a^n<(a+h)^n<a^n+2nha^{n−1}; \text{for sufficiently small $h>0$ and $n>0$}$$ Also, there is no condition on $a$ or $n$, but if $a<0$, isn't the expression $a^n$ undefined for irrational values of $n$?

Thus, I want to know the proofs of both the statements given above, and any resources for further learning in these topics(Advanced limit proofs to be specific, as I haven't seen the proof of the first statement in any standard real analysis book so far, the case for integral values of $n$ is proved, and the author mostly asks us to accept the statement for reall values of $n$ as well..).

Thanks for any answers!!

thornsword
  • 687
  • For irrational $n$ you need $a>0$ and some definition of symbol $a^n$ (eg $a^n=\exp(n\log a) $). For rational $n$ a more general result is proved here. – Paramanand Singh May 09 '20 at 08:41
  • Factorize $(a+h)^n-a^n$. – Aravind May 09 '20 at 08:46
  • Neither the linked question, nor any of the answers include "$\forall n \in \Bbb R$". That was your addendum. Since $n$ is usually used for integers, the answers there assume $n$ is integer. In fact, the accepted answer says to use induction on $n$ (which requires $n$ to be integer). Since the asker accepted that answer, this was apparently their intent as well. When $a < 0$ and $n$ is not integer, neither $x^n$ nor $a^n$ is well-defined by convention. You can choose a definition for them (allowing complex values), but then the result will depend on how you choose to define them. – Paul Sinclair May 09 '20 at 20:53

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