Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some constants.
I know how to calculate this limit using the L'Hopital's rule:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$
$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$
$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot 
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$
$$
\Rightarrow 
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$
I'm allowed to use the limits $\lim_{x\to0}\frac{\sin x}{x}=1$,
$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$,
$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$ and $\lim_{x\to0} (1+x)^{1/x}=e$.

Let's prove this expression:
Given positive $a$ and $b$:
$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}$$.
In order to show this we are going to use the following:
$$\lim_{n\rightarrow\infty}n(\sqrt[n]{a}-1)=\ln a$$ thus by AM-GM
$$\frac{1}{n}\ln\sqrt{ab}\leq\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})=\ln\bigg(\frac{1}{2}(\sqrt[n]{a}-1)+\frac{1}{2}(\sqrt[n]{b}-1)+1\bigg)<\frac{1}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$
now multiply both sides by $n$, we get
$$\ln\sqrt{ab}\leq n\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})\leq\frac{n}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$
Taking $n\rightarrow\infty$ and using the squeeze theorem:
We get
$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}.$$
Let $n=\frac1x$ and the rest is done.
