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So, here is the clear formulation of the problem: let $(x_n) $ be a convergent sequence of positive numbers, with $x_n \to c$. I want to prove that the sequence $(y_n) $, with $y_n=a^{x_n} $, tends to $a^c$

I will use the proof by contradiction. Suppose that $y_n \to, l$ with $l \neq a^c$.

But, we know that if $y_n \to l$, then $\ln{y_n} \to \ln l$. That means that:

$$\lim{\ln{y_n}} =\ln l$$ $$\lim{\ln{a^{x_n}} } =\ln l$$ $$\lim{(x_n\ln{a}) } =\ln l$$ $$\ln{a}(\lim{x_n}) =\ln l$$ $$c\ln{a} =\ln l$$ $$\ln{a^c} =\ln l$$

And now, from the injectivity of logarithm function it follows that $l=a^c$, which is a contradiction. Hence, Q. E. D.

Please, can you tell me if my proof is a correct one?

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  • $\begingroup$ In the step where you "push" the limit under the logarithm, you implicitly invoke continuity of logarithm. The proof is ok, though this approach is a bit more complicated than using continuity directly to conclude that $\ln a^{x_n}\to\ln a^c$. $\endgroup$ – Marcin Łoś May 18 '14 at 18:23
  • $\begingroup$ Well, in fact, I don't have the concept of continuity very clear in mind, I haven't studied it yet. Although, it seems that here math.stackexchange.com/questions/800534/… I proved the continuity of the ln function. How would a shorter proof look? Can yoy show me, please? $\endgroup$ – Bardo May 18 '14 at 18:38
  • $\begingroup$ The setup of your proof is invalid. You would need to prove that $y_n$ converges at all. $y_n \not \to a^c$ means $y_n \to l \neq a^c$ or $y_n$ diverges. $\endgroup$ – asmeurer May 18 '14 at 18:45
  • $\begingroup$ So I need to erase the part where I state that I use proof by contradiction, right? $\endgroup$ – Bardo May 18 '14 at 18:55
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This is correct provided you're allowed to assume that $\lim (\ln y_n) = \ln (\lim y_n)$.

Your proof structure could be simplified, though: you gave a self-contained proof that $l=a^c$, but surrounded it unnecessarily with 'suppose $l \ne a^c$' and 'contradiction'. If you just didn't assume the result was false (and therefore never reached a contradiction) then you'd still have a proof.

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  • $\begingroup$ Wait, that's right. Yes, if l=a^c, then that's it, because the limit of a sequence is unique.. $\endgroup$ – Bardo May 18 '14 at 18:34

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