In the Mean Value Theorem, $f(x+h)=f(x)+hf'(x+ \theta h)$ where $0< \theta <1$, $f(x)=\sin x$ In the Mean Value Theorem, $f(x+h)=f(x)+hf'(x+ \theta h)$ where $0< \theta <1$, prove that $\lim_{h \rightarrow 0} \theta=\frac{1}{2}$, if $f(x)=\sin x$
After a chain of obvious calculations, I am getting $\lim_{h \rightarrow 0} \theta=\lim_{h \rightarrow 0} \frac{\arccos(\frac{\sin(x+h)-\sin x}{h})-x}{h}$
How to show that this is $\frac{1}{2}$?
 A: We know that $\lim_{h\rightarrow\infty}\frac{\sin(x+h)-\sin x}{h}=\cos x$. Hence the numerator $\left[\arccos\left(\frac{\sin(x+h)-\sin x}{h}\right)-x\right]$ tends to $0$ as $h\rightarrow 0$. So apply L'Hospital's rule.
A: A solution without l'Hospital's rule:
Consider $$g(h) = \begin{cases} \frac{\sin(x+h)-\sin x}{h} & h\not=0 \\ \cos(x) & h=0\end{cases}$$
Then we have by using the mean value theorem for a $\xi \in (0,h)$
$$\lim_{h \rightarrow 0} \frac{\arccos(\frac{\sin(x+h)-\sin x}{h})-x}{h} = \lim_{h \rightarrow 0} \frac{\arccos(g(h)) - \arccos(g(0))}{h - 0} = \lim_{h \to 0} \left(\arccos g(\xi)\right)'$$
And because with $h \to 0$ we have $\xi \to 0$ we get
$$\lim_{h \rightarrow 0} \frac{\arccos(\frac{\sin(x+h)-\sin x}{h})-x}{h} =\lim_{\xi \to 0} \left(\arccos g(\xi)\right)'$$
A: In general, if $f$ is sufficiently smooth, you get
\begin{align}
&f(x+h)&&=f(x)+f'(x)h+\frac12f''(x)h^2+O(h^3)
\\
&=f(x)+f'(x+θ_hh)h&&=f(x)+f'(x)h+f''(x)θ_hh^2+O(h^3)
\end{align}
so that in all situations where $f''(x)\ne 0$ you get $$θ_h=\frac12+O(h).$$

For $f(x)=\sin(x)$ consider the identified critical point $x=0$. There we get $$
\sin(0+h)=\sin(0)+\cos(0+θ_hh)h=\frac{\sin h}{h}\,h
$$ 
so that 
$$
\cos(θ_hh)=\frac{\sin h}{h}\\
1-\frac12(θ_hh)^2+O(h^4)=1-\frac16h^2+O(h^4)\\
θ_h=\frac1{\sqrt3}+O(h^2)
$$
 which contradicts the claim.
