Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives

Question

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 $\lim_{x\to0} (1+x)^{1/x}=e$.

Answer 1

This post has multiple answers that explain why, if $\displaystyle\lim_{x\to a} f(x)=1$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ then $$\lim_{x\to a} (f(x))^{g(x)}=e^{\lim\limits_{x\to a} (f(x)-1)g(x)}$$ Using this formula, calling the limit as L, we have $$L=e^{\lim\limits_{x\to 0}\left(\frac{a^{\sin x}+b^{\sin x}-2}{2x}\right)}$$ so $$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2x}+ \dfrac{b^{\sin x}-1}{2x}\right)$$$$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2\sin x}\cdot\frac{\sin x}{x}+ \dfrac{b^{\sin x}-1}{2\sin x} \frac{\sin x}{x}\right)$$$$=\frac{\ln a}{2}+\frac{\ln b}{2}=\frac{\ln ab}{2}=\ln\sqrt {ab}$$ whence $L=\sqrt{ab}$