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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.

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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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I see how you got $lim_{n \rightarrow \infty} a ^{\r{n}} = a^{-x} > 0$ but I don't know why it implies the second part thanks. –  Jmaff Dec 26 '12 at 22:58
    
Take the multiplicative inverse of both sides –  Amr Dec 26 '12 at 23:21
    
hmm, still unsure. If I take the inverse of the first line then we get $(lim_{n \rightarrow \infty} a^{r_{n}}) \cdot a^{x} =1$ Or should I look at the inverses of the actual sequence terms? –  Jmaff Dec 26 '12 at 23:34
    
The inverses of the actual terms of the sequence (you can take their inverse becasue they are non zero) –  Amr Dec 26 '12 at 23:35
    
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,Thanks –  Jmaff Dec 26 '12 at 23:43

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 $\frac{\lim_n a^{r_n} }{a^x}$

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