Prove the following limit: \lim_{x \to 0}\left(\frac{a^x+b^x}{2}\right)^{3/x}=ab\sqrt{ab}
Solution:
This is one of the limit problems I consider relatively difficult. From
the Maclaurin series, we can derive: a^x=1+x
\ln a+\frac{x^2 \ln^2 a}{2!}+\frac{x^3 \ln^3 a}{3!}+\dots
Thus, the original expression can be transformed into: \lim_{x \to 0}\left(\frac{2+x \ln a+\frac{x^2 \ln^2 a}{2!}+\dots+x \ln b+\frac{x^2 \ln^2 b}{2!}+\dots}{2}\right)^{3/x}
We have a simple limit: \lim\limits_{x\to 0}(a+x^2)^{1/x}=a^{1/x}. Therefore, in the above expression, the terms \frac{x^2 \ln^2 a}{2!} and subsequent terms can be ignored, considering only: \begin{aligned} &\,\lim_{x \to 0}\left(\frac{2+x \ln a+x \ln b}{2}\right)^{3/x}\\ =&\,\lim_{x \to 0} \left\{\left[1+\left(\frac{\ln a+\ln b}{2}\right)x\right]^{1/x}\right\}^3\\ =&\,e^{\frac{3(\ln a+ \ln b)}{2}}\\ =&\,(ab)^{3/2}\\ =&\,ab\sqrt{ab} \end{aligned}
Similarly, we have: \lim_{x \to 0}\left(\frac{a_1^x+a_2^x+\dots+a_n^x}{n}\right)^{1/x}=\sqrt[n]{a_1 a_2 \dots a_n}
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