Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
You didn't mean $\ln$, you mean exp$()$ –  Varun Iyer Aug 10 at 16:39
    
$a^{x}=e^{xln(a)}$ –  Maman Aug 10 at 16:43
    
apply l'hopital rule –  user45765 Aug 10 at 16:44
1  
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. –  varun Aug 10 at 16:49
1  
I'm truely sorry for the typo. Corrected –  Elimination Aug 10 at 16:56

2 Answers 2

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 $$ \lim_{x\to-\infty}\mu_{a,b}(x)=\min(a,b),\qquad \lim_{x\to\infty}\mu_{a,b}(x)=\max(a,b). $$

share|improve this answer
    
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 at 17:05
    
@QuangHoang Yes, of course. –  egreg Aug 10 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 $\frac{\ln ab}{2}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.