A Limit Proof Problem Involving a and b

关于a,b的极限证明题目 苏剑林 · 2009-08-24 · Mathematics proof limit

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 fairly difficult. From Maclaurin's formula we can derive:

$$a^x=1+x \ln a+\frac{x^2 \ln^2 a}{2!}+\frac{x^3 \ln^3 a}{3!}+...$$more

so the original expression can be turned into

$$\lim_{x \to 0}\left(\frac{2+x \ln a+\frac{x^2 \ln^2 a}{2!}+...+x \ln b+\frac{x^2 \ln^2 b}{2!}+...}{2}\right)^{3/x}$$

We have a simple limit: $\lim\limits_{x\to 0}(a+x^2)^{1/x}=a^{1/x}$, and therefore, in the expression above, $\frac{x^2 \ln^2 a}{2!}$ and the terms following it can be neglected, leaving us to consider only

$$\begin{aligned} &\,\lim_{x \to 0}\left(\frac{2+x \ln a+x \ln b}{2}\right)^{3/x}\\ =&\,\lim_{x \to 0} \left\{\left[1+\left(\frac{\ln a+\ln b}{2}\right)x\right]^{1/x}\right\}^3\\ =&\,e^{\frac{3(\ln a+ \ln b)}{2}}\\ =&\,(ab)^{3/2}\\ =&\,ab\sqrt{ab} \end{aligned}$$

By the same reasoning, we have

$$\lim_{x \to 0}\left(\frac{a_1^x+a_2^x+...+a_n^x}{n}\right)^{1/x}=\sqrt[n]{a_1 a_2...a_n}$$