TITLE: Cute limit: $\lim\limits_{x \to 0^{+}}\frac{\sin(x)^x − x^ {\sin(x)}}{\tan(x) ^x − x^{\tan (x)}}$
QUESTION [6 upvotes]: What's the fastest way to calculate the below limit (ideally, without resorting to Taylor expansions)?
$$\lim_{x \to 0^{+}}\frac{\sin(x)^x
 − x^ {\sin(x)}}{\tan(x)
^x − x^{\tan (x)}}$$

REPLY [8 votes]: The Mean Value Theorem says
$$
\frac{e^x-e^y}{x-y}=e^\xi
$$
for some $\xi$ between $x$ and $y$. Therefore,
$$
\begin{align}
\lim_{x\to0^+}\frac{\sin(x)^x−x^{\sin(x)}}{\tan(x)^x−x^{\tan(x)}}
&=\lim_{x\to0^+}\frac{x\log(\sin(x))-\sin(x)\log(x)}{x\log(\tan(x))-\tan(x)\log(x)}\tag{1}\\[6pt]
&=\lim_{x\to0^+}\frac{x\log(x)+x\log\left(\frac{\sin(x)}{x}\right)-\sin(x)\log(x)}{x\log(x)+x\log\left(\frac{\tan(x)}{x}\right)-\tan(x)\log(x)}\tag{2}\\
&=\lim_{x\to0^+}\frac{x-\sin(x)+\frac{x}{\log(x)}\log\left(\frac{\sin(x)}{x}\right)}{x-\tan(x)+\frac{x}{\log(x)}\log\left(\frac{\tan(x)}{x}\right)}\tag{3}\\
&=\lim_{x\to0^+}\frac{x-\sin(x)+\frac{x}{\log(x)}O\!\left(\frac{x-\sin(x)}{x}\right)}{x-\tan(x)+\frac{x}{\log(x)}O\!\left(\frac{x-\tan(x)}{x}\right)}\tag{4}\\
&=\lim_{x\to0^+}\frac{x-\sin(x)+O\!\left(\frac{x-\sin(x)}{\log(x)}\right)}{x-\tan(x)+O\!\left(\frac{x-\tan(x)}{\log(x)}\right)}\tag{5}\\[9pt]
&=\lim_{x\to0^+}\frac{x-\sin(x)}{x-\tan(x)}\tag{6}\\[15pt]
&=\lim_{x\to0^+}\frac{1-\cos(x)}{1-\sec^2(x)}\tag{7}\\[15pt]
&=\lim_{x\to0^+}-\frac{\cos^2(x)}{1+\cos(x)}\tag{8}\\[15pt]
&=-\frac12\tag{9}
\end{align}
$$
Explanation:
$(1)$: Mean Value Theorem
$(2)$: $\log(ab)=\log(a)+\log(b)$
$(3)$: divide numerator and denominator by $\log(x)$
$(4)$: $\log(1+x)=O(x)$
$(5)$: algebra
$(6)$: $O\!\left({\raise{1.5pt}\frac{u}{\log(x)}}\right)=o(u)$ as $x\to0^+$
$(7)$: L'Hôpital
$(8)$: multiply numerator and denominator by $\frac{\cos^2(x)}{1-\cos(x)}$
$(9)$: evaluate at $x=0$