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I have to calculate the following limit: $$\lim_{x \to 0} \left[\frac{a_1^x+a_2^x+\cdots+a_n^x}{n}\right]^{\frac{1}{x}}$$ I said that as $x→0$ the $a_1^x+a_2^x+\cdots+a_n^x$ are approaching $n$ since we have $n$ terms, so we will get: $$\lim_{x \to 0} \left[\frac{n}{n}\right]^{\frac{1}{x}}$$ $$\therefore \lim_{x \to 0} \left(1\right)^{\frac{1}{x}}$$ as $x→0$, $1/x→\infty$ $$\therefore \left(1\right)^{\infty}=1$$ But one friend said that it was a false way to calculate that limit, where is my error and what is it's solution?

(where $a_i$ are positive real numbers)

Did
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    If $a_1 = a_2 = \ldots = a_n = \alpha$, then $$\lim_{x\to 0} \left [ \frac{a_1^x + a_2^x + \ldots + a_n^x}{n}\right ]^{1/x} = \alpha$$ So this begs the question : What are the $a_i$'s? – Prahlad Vaidyanathan Nov 18 '13 at 14:07
  • While the term inside the parenthesis approaches $1$ for $x \to 0$, the exponent $\frac1x$ approaches $\infty$. Raising a number close to $1$ to a large power can produce a number that is no longer close to $1$. – Daniel Fischer Nov 18 '13 at 14:08
  • Related: http://math.stackexchange.com/questions/487250/how-find-this-nice-limt but not a duplicate (though there are duplicates, I just haven't found them yet). – Daniel Fischer Nov 18 '13 at 14:11

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