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For which values of $a$ the following limit is true?

$$x,a\in \mathbb R$$ $$\displaystyle\lim_{x \to -\infty} x^a = -\infty$$

For whole even numbers that would be false, and for whole odd numbers, that would be true.

What about the rest?

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  • $\begingroup$ Your results so far are corretcd only for $a>0$. You can easily solve the remaining cases $a\in\mathbb Z\setminus\mathbb N$. How do you define $x^a$ for $x<0$ and $a\notin \mathbb Z$? $\endgroup$ Aug 6, 2014 at 17:06
  • $\begingroup$ what do you mean how do i define, isnt $x^a$ well defined for every $a\in \mathbb R$? $\endgroup$
    – guynaa
    Aug 6, 2014 at 17:08
  • $\begingroup$ What is $(-1)^{1/2}$? $(-1)^\pi$? $\endgroup$ Aug 6, 2014 at 17:09
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    $\begingroup$ What is $(-\pi)^\pi$? $\endgroup$ Aug 6, 2014 at 17:09
  • $\begingroup$ how about defining it as the limit of $x^(a_n)$ such that $a_n\in \mathbb Q$ , $a_n \to a$ $\endgroup$
    – guynaa
    Aug 6, 2014 at 17:12

2 Answers 2

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Real exponents with negative bases is actually not well-defined. You can read this short explanation:

http://en.wikipedia.org/wiki/Exponentiation#Real_exponents_with_negative_bases

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Suppose $a=j/k$ is a fraction in lowest terms and all numbers under consideration are assumed to be real. If both $j$ and $k$ are odd then $\lim_{x \to -\infty} =-\infty$. If $j$ is even then $k$ is odd and so $\lim_{x \to -\infty} =\infty$. If $j$ is odd and $k$ is even then the limit doesn't exist. If $a$ is an irrational number then in any neighborhood of $a$ there are fractions whose denominators are even and so the limit does not exist. This last part assumes that we are defining an irrational number using limits of sequences of rational numbers. See http://en.wikipedia.org/wiki/Dedekind_cut

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