Originally Posted by

**galactus** I like seeing others do these, but I thought I would throw my hat in.

I will take a stab at one. But I like to watch mathstud, Moo, and others go at them. Then I can absorb some of their 'useful tricks'.

Here is a method I used for this particular integral some years back. I have seen it before.

It just entails a little algebra and e.

Here is #4 using a little algebra and a substitution.

$\displaystyle \lim_{x\to 0}\left(\frac{a^{x}+b^{x}}{2}\right)^{\frac{1}{x}}$

Let $\displaystyle t=\frac{1}{x}, \;\ x=\frac{1}{t}$

Then, we note that as $\displaystyle {x\to 0}$, $\displaystyle t\to {\infty}$

Make the sub and we get:

$\displaystyle \lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}\right)^{t}$

Now, a little algebra and 'e'; add 1 and subtract 1:

$\displaystyle \lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}+1-1\right)$

$\displaystyle \lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}-2}{2}+1\right)^{t}$

Because of asymptotics that +1 is just a 'hanger on' and we can rewrite as such using e:

$\displaystyle \lim_{t\to {\infty}}e^{t\left(\frac{a^{\frac{1}{t}}+b^{\frac{ 1}{t}}-2}{2}\right)}$

$\displaystyle \lim_{t\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}} +b^{\frac{1}{t}}-2}{\frac{1}{t}}\right)}$

$\displaystyle \lim_{n\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)}$

Now, the limit, which is the exponent of e, can be taken:

$\displaystyle \frac{1}{2}\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)$

Now, the limits here I won't bother deriving. I will use it as a 'lemma'

$\displaystyle \lim_{t\to {\infty}}(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}})=\lim_{t\to {\infty}}(a^{\frac{1}{t}}-1)t=ln(a)$

Analogous with b, which is ln(b).

So, we end up with:

$\displaystyle e^{\frac{1}{2}(ln(a)+ln(b))}$

$\displaystyle e^{ln(\sqrt{ab})}$

Which leads to $\displaystyle \sqrt{ab}$