# Evaluating a limit. What makes the equality right?

I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$ where $a,b>0$.

The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} = \exp\left( \lim\limits_{x\to 0}\frac{\frac{a^x+b^x}{2} - 1}{x} \right)$$

How come?

Update:
Of course, $$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} = \exp\left(\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^x+b^x}{2} \right)}{x} \right)$$

But how to proceed to reach the auther's expression?

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You didn't mean $\ln$, you mean exp$()$ – Varun Iyer Aug 10 '14 at 16:39
$a^{x}=e^{xln(a)}$ – Maman Aug 10 '14 at 16:43
apply l'hopital rule – user45765 Aug 10 '14 at 16:44
You can use $$\lim_{y\rightarrow 0} \frac{\ln (1+y)}{y}=1$$ with $y=\frac{a^x+b^x}{2}-1$. Surely there is some typo with the claim as the limit inside the RHS exponential is blowing up. – theindigamer Aug 10 '14 at 16:49
I'm truely sorry for the typo. Corrected – Elimination Aug 10 '14 at 16:56

If we try with $$\lim_{x\to 0} \frac{\log(a^x+b^x)-\log 2}{x}$$ and apply l'Hôpital's theorem, we get $$\lim_{x\to 0}\frac{a^x\log a+b^x\log b}{a^x+b^x}=\frac{\log a+\log b}{2}= \log\sqrt{ab}.$$ It's just the derivative of $x\mapsto (a^x+b^x)/2$ at $0$, of course.

However, $$\lim_{x\to0}\frac{\log(1+x)}{x}=1$$ so that $$\lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{x}= \lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{\dfrac{a^x+b^x}{2}-1} \frac{\dfrac{a^x+b^x}{2}-1}{x}$$ and the limit of the first factor is $1$. I don't think it's a real simplification.

It may be worth noting that the function $$\mu_{a,b}(x)=\begin{cases} \left(\dfrac{a^x+b^x}{2}\right)^{1/x} & \text{if x\ne0}\\[2ex] \sqrt{ab} & \text{if x=0} \end{cases}$$ for $a,b>0$ is quite interesting, because it's increasing, $\mu_{a,b}(-1)$ is the harmonic mean, $\mu_{a,b}(0)$ is the geometric mean, $\mu_{a,b}(1)$ is the arithmetic mean and $$\lim_{x\to-\infty}\mu_{a,b}(x)=\min(a,b),\qquad \lim_{x\to\infty}\mu_{a,b}(x)=\max(a,b).$$

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A little off topic, but when you use l'Hopital, you can just say that it's the derivative of $\log(a^x+b^x)$ at $x=0$. – Quang Hoang Aug 10 '14 at 17:05
@QuangHoang Yes, of course. – egreg Aug 10 '14 at 17:07

First use simple fact, that $\displaystyle\lim_{y \to 0}\frac{\ln(1+y)}{y}=1$, so:

$$\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{y}=\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{\frac{a^x+b^x}{2}-1} \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{y}=\\=1 \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}$$

Now $\lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}=\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}+\lim_{x \to 0}\frac{1}{2}\frac{b^x-1}{x}$

But $a^x=e^{x \ln a }$, so $\lim_{x \to 0}\frac{1}{2}\frac{a^x-1}{x}=\lim_{x \to 0}\ln a\frac{1}{2}\frac{e^{\ln a x}-1}{x \ln a}=\frac{1}{2}\ln a$. The same with second limit. Finally the result is $\frac{\ln ab}{2}$.

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