Sorry if I make something wrong in this post, because it is my first post.
$$ \lim_{x\rightarrow 0}\left [ \frac{a_{1}^{x}+a_{2}^{x}+a_{3}^{x}+\dots+a_{n}^{x}}{n} \right ]^{1/x} $$
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What are your thoughts on this problem? – abiessu Apr 29 '14 at 21:17
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@abiessu I'm trying to solve this by number "e", but I can't. What are your suggesions, any method? – ExpikQwg Apr 29 '14 at 21:20
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You must be assuming all $a_k \geq 0$? – user2566092 Apr 29 '14 at 21:21
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http://math.stackexchange.com/questions/531831/why-is-the-0th-power-mean-defined-to-be-the-geometric-mean – Apr 29 '14 at 21:22
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Are there any rrstrictions about $x$ and $a_j$? – ajotatxe Apr 29 '14 at 21:25
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@ByronSchmuland is there any way you know expect assuming? By the way, thank you for your suggestions. – ExpikQwg Apr 29 '14 at 21:27
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@ExpikQwg I don't understand the sentence "is there any way you know expect assuming?" – Apr 29 '14 at 21:32
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@ByronSchmuland do you know any other way to solve this limit expect this one with L'Hospital's rule (link you posted). Thank you. – ExpikQwg Apr 29 '14 at 21:34
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@ExpikQwg You want the word "except", not "expect". I'm sorry, I don't know any other way. – Apr 29 '14 at 21:35
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@ByronSchmuland sorry for that mistake :) – ExpikQwg Apr 29 '14 at 21:37
1 Answers
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"Shortcut":
If
$\lim_{x→α}f(x)=1$ and $\lim_{x→α}g(x)= \infty $
then $$\lim_{x→α}(f(x))^{g(x)}= \lim_{x→α}(1+f(x)−1)^{g(x)}= \lim_{x→α}[[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{(f(x)-1)}]^{g(x)}= \lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$ $$\lim_{x\rightarrow 0}\left [ \frac{a_{1}^{x}+a_{2}^{x}+a_{3}^{x}+\dots+a_{n}^{x}}{n} \right ]^{1/x}=e^{\frac{1}{n}\lim_{x\rightarrow 0}(\dfrac{a_{1}^{x}-1}{x}+\dfrac{a_{2}^{x}-1}{x}+ \cdots +\dfrac{a_{n}^{x}-1}{x})}=e^{\frac{1}{n}(\ln a_1+\ln a_2+ \cdots +\ln a_n)}= e^{\ln (a_1\cdot a_2\cdots a_n)^{\frac{1}{n}}}= (a_1\cdot a_2\cdots a_n)^{\frac{1}{n}}.$$
We applied $\lim_{x\rightarrow 0}\dfrac{a^{x}-1}{x}= \ln a.$
medicu
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