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Let $a>0$, and let $x$ be a real number. Prove that if $\{r_{n} \}$ is any decreasing rational sequence with limit $x$, then $a^x = \lim_{n \rightarrow \infty} a^{r_n}$

Where in the book $a^x$ is defined as $\lim_{n \rightarrow \infty} a^{s_n}$ where $ \{ s_n \}$ is an increasing sequence.

2 Answers 2

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Note that $\{-r_n\}$ is an increasing sequence of rationals that converges to $-x$

Using the definition of your book, we know that: $lim_{n\rightarrow\infty}a^{-r_n}=a^{-x}>0$

Since the limit is positive, thus: $lim_{n\rightarrow\infty}a^{r_n}=a^{x}$

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    Okay,so since we know that the $\{ -r_n \}$ sequence has an invertible limit, and that each element of the sequence is invertible we know that the limit of the sequence of the inverses has the limit that is the inverse of$\{ -r_n \}$'s limit ? Is this a commonly proved statement? I think it can be prove using the limit rule for limits maybe,Thanks2012-12-26
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Since $2x-r_n$ is increasing and

$\lim_n 2x-r_n=x$

we have

$a^x=\lim_n a^{2x-r_n}=\lim_n \frac{a^{2x}}{a^{r_n}}= \frac{a^{2x}}{\lim_n a^{r_n}} \,.$

Multiply both sides by $\frac{\lim_n a^{r_n} }{a^x}$