Evaluate the limit with exponents using L'Hôpital's rule or series expansion

Question

Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}} -\sqrt{ab}}{x}$$ It is known that $a>0,b>0$

My Attempt:

I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$

Answer 1

We have that

• $a^x=e^{x\log a}=1+x\log a+\frac12x^2\log^2 a+o(x^2)$

• $b^x=e^{x\log b}=1+x\log b+\frac12x^2\log^2 b+o(x^2)$

then

$$\left( \frac{a^x+b^x}{2} \right)^{\frac{1}{x}}=\left(\frac{2+x\log ab+\frac12x^2(\log^2 a+\log^2 b)+o(x^2)}{2}\right)^{\frac{1}{x}}=\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)^{\frac{1}{x}}=e^{\frac{\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)}{x}}$$

and since

$$\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)\\=x\log \sqrt{ab} +\frac14x^2(\log^2 a+\log^2 b)-\frac12x^2\log^2\sqrt{ab}+o(x^2)$$

we have that

$$e^{\frac{\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)}{x}}=\sqrt {ab}(1+\frac14x(\log^2 a+\log^2 b)-\frac12x\log^2\sqrt{ab}+o(x))$$

therefore

$$\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}} -\sqrt{ab}}{x}= \sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac12\log^2\sqrt{ab}+o(1)\right)\to \sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac12\log^2\sqrt{ab}\right) =\sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac18(\log a+\log b)^2\right)=\frac{\sqrt {ab}}8\left(\log a-\log b\right)^2$$

Answer 2

Not sure how you could fathom that

But after that the limit becomes $$\lim_{h\to0}\dfrac{f(0+h)-f(0)}{h-0}=f'(x)_{\text{ at }x=0}$$ where $f(x)=\left(\dfrac{a^x+b^x}2\right)^{1/x}$

Answer 3

just use the l-H hospital for it,a tip for that is you can write the $((a^x+b^x)/2)^{1/x}$ to this inform: $e^{[\ln((a^x+b^x)/2)]/x}$,so you can use chain rule to differeitiate it,here you are,the answer leaves for you

Answer 4

Using the series expansion for $\log(1+x)$, we get $$ \frac1x\,\log\left(1+mx+nx^2+O\!\left(x^3\right)\right) =m+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right) $$ Using the series expansion for $e^x$, we get $$ \left(1+mx+nx^2+O\!\left(x^3\right)\right)^{1/x} =e^m\left(1+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right)\right) $$ Therefore, writing $a^x=e^{x\log(a)}$ and $b^x=e^{x\log(b)}$ and using the series expansion for $e^x$, we get $$ \begin{align} \frac{\left(\frac{a^x+b^x}2\right)^{1/x}-\sqrt{ab}}x &=\frac{\scriptsize\left(1+\frac x2(\log(a)+\log(b))+\frac{x^2}4\left(\log(a)^2+\log(b)^2\right)+O\!\left(x^3\right)\right)^{1/x}-\sqrt{ab}}x\\ &=\frac{\sqrt{ab}\left(1+\frac{\log(a)^2+\log(b)^2-2\log(a)\log(b)}8x+O\!\left(x^2\right)\right)-\sqrt{ab}}x\\[9pt] &=\frac18\sqrt{ab}\,(\log(a)-\log(b))^2+O(x) \end{align} $$ Thus, $$ \lim_{x\to0}\frac{\left(\frac{a^x+b^x}2\right)^{1/x}-\sqrt{ab}}x =\frac18\sqrt{ab}\,(\log(a)-\log(b))^2 $$