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I found this problem interesting to be here:

If $a,b>0$ then find the following limit: $$\lim_{x\to\pm\infty}\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x$$

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It's bounded from above by $(a+b)/2$ –  akkkk Dec 23 '12 at 18:44
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The answer is the geometric mean, and is direct (from the AM-GM inequality) if we assume the limits for $\infty$ and $-\infty$ are equal. –  ronno Dec 23 '12 at 18:46
    
A related question. –  user26872 Dec 23 '12 at 19:01

1 Answer 1

up vote 6 down vote accepted

One can use L'Hospital's Rule mechanically. The logarithm of our expression is $$\frac{\log\left(\frac{a^{1/x}+b^{1/x}}{2}\right)}{1/x}.$$ Differentiate top and bottom. We get $$\frac{1}{\frac{a^{1/x}+b^{1/x}}{2}}(-1/x^2)\frac{\frac{a^{1/x}\log a+b^{1/x}\log b}{2}}{(-1/x^2)}$$ Cancel the $(-1/x^2)$. The limit is $\dfrac{\log a+\log b}{2}$. Then exponentiate.

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