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This confused me,

$$\lim_{n \to \infty} x^{1 + \frac{1}{2n-1}} = x \lim_{n \to \infty}x^{\frac{1}{2n-1}}=|x| $$

Which means $$\lim_{n \to \infty}x^{\frac{1}{n}} = \left\{ \begin{array}{lr} 1 & : x > 0\\ -1 & : x < 0\\ 0 & : x = 0 \end{array} \right.$$

Rudin gives an easy proof as to why this works when x is nonnegative, but when x is negative I don't know how to deal with fractional powers of negative numbers. Is this something I simply can't do without further reading or am I being stupid?

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Jonas Meyer: I can see why you would think so. I made a jump in reasoning there and Raymond explains why it was incorrect. –  StuartHa Jan 10 '12 at 20:12
    
Fixed. Thanks.. –  StuartHa Jan 10 '12 at 20:49

1 Answer 1

up vote 5 down vote accepted

Note that Rudin was cautious since he had $\lim_{n \to \infty}x^{\frac{1}{2n-1}}$ so that he always considered a $(2n-1)$th (odd!) root of x.

Your inference is wrong : this doesn't mean that $\lim_{n \to \infty}x^{\frac{1}{n}}$ is $1$ or $-1$ depending of the sign of $x$. If $x\lt0$ this last limit doesn't exist because the (2n)-th root is not defined in $\mathbb{R}$.

I think that Rudin's point was that $$\lim_{n \to \infty}x^{\frac{1}{2n-1}} = \left\{ \begin{array}{lr} 1 & : x > 0\\ -1 & : x < 0\\ 0 & : x = 0 \end{array} \right.$$

which is true (keeping the parity of n is vital!).

To understand what's going on just consider an example : $a_n=(-2)^{\frac{1}{2n-1}}$ you'll get a number closer and closer to $-1$ as $n \to \infty$.

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