Where power means lead

Idea. For $x=1$ the expression is the arithmetic mean $\frac{a+b}2$. For $x=-1$ it is related to the harmonic mean. What happens when $x\to0$? Let us compute.

Step 1: an expansion. Since $\dfrac{e^{x\ln a}-1}{x}\to\ln a$, we have $a^x=1+x\ln a+o(x)$, and similarly for $b^x$. Hence
$$\frac{a^x+b^x}{2}=1+h,\qquad h=x\,\frac{\ln a+\ln b}{2}+o(x)=x\ln\sqrt{ab}+o(x).$$

Step 2: the logarithm. As $h\to0$, $\ln(1+h)=h+o(h)=x\ln\sqrt{ab}+o(x)$. Therefore
$$\frac1x\ln\!\left(\frac{a^x+b^x}2\right)\longrightarrow\ln\sqrt{ab}.$$

Step 3: the exponential. By continuity of $\exp$, the original expression tends to $e^{\ln\sqrt{ab}}=\sqrt{ab}$.

Conclusion. The limit is the geometric mean $\sqrt{ab}$.

A numerical check with $a=1$, $b=4$ (so $\sqrt{ab}=2$):

• $x=1$: $2.5$
• $x=0.1$: $2.0486$
• $x=0.01$: $2.0048$
• $x=0.001$: $2.0005$

The values slide from the arithmetic mean $2.5$ towards the geometric mean $2$.