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 $\lim_{x\to0} (1+x)^{1/x}=e$.
 A: 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 $L=\sqrt{ab}$
A: Taking logarithm, then using $\sin x\sim x$ and $\ln(1+x)\sim x$ when $x\to0$, we have
\begin{align*}
\frac1x\cdot\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)&\sim\frac1{\sin x}\cdot\left(\frac{a^{\sin x}+b^{\sin x} }{2}-1\right)\\
&=\frac12\frac{a^{\sin x}-1}{\sin x}+\frac12\frac{b^{\sin x}-1}{\sin x}\\
&\rightarrow\frac12(\ln a+\ln b),\qquad x\to0,
\end{align*}
where in the last line we've used
$$\lim_{x\to0}\frac{a^{\sin x}-1}{\sin x}=\lim_{t\to0}\frac{a^{t}-1}{t}=\ln a. $$
Therefore,
$$\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.$$
A: 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.
