Finding limit at $1/2$ not using L'Hôpital's rule Find
$$
\lim_{x \to 1/2}\left[%
\frac{\arcsin\left(x\right) - \arcsin\left(1/2\right)}{x - 1/2}\right]
$$
I've gotten the answer to be $2/\sqrt{3}$ due to L'Hôpital's rule but I'm trying to find it using an elementary way (non-L'Hôpital method) and don't seem to have much luck :(
Help ?
 A: So that there won't be complains:
$$\lim_{x\to 1/2}\frac{\arcsin x-\arcsin\frac12}{x-\frac12}=:\left(\arcsin\frac12\right)'=\frac1{\sqrt{1-\frac12}}$$
A: An alternative way:
The statement is that $\arcsin'(\frac12)=2/\sqrt3$. Finding the derivative of $\arcsin$ can be done as follows:
From the definition of $\arcsin$ we have:
$$\arcsin(\sin(x))=x$$
Deriving both sides gives:
$$\cos(x) \arcsin'(\sin(x))=1\\
\arcsin'(\sin(x))=\frac1{\cos(x)}$$
Using trigonometric identifies gives:
$$\arcsin'(\sin(x))=\frac1{\sqrt{1-\sin^2x}}$$
Which means:
$$\arcsin'(x)=\frac1{\sqrt{1-x^2}}$$
Then
$$\lim_{x\to\frac{1}{2}}\frac{\arcsin x-\arcsin\frac{1}{2}}{x-\frac{1}{2}}=\arcsin'(\frac12)=\frac1{\sqrt{1-(\frac12)^2}}=\frac1{\sqrt{\frac34}}=\frac2{\sqrt3}$$
A: This is computing the derivative of $\arcsin$ at $1/2$. It's easier if you invert your fraction and set $\arcsin x=y$.
If $x\to\frac{1}{2}$, then $y\to\frac{\pi}{6}$, so
$$
\lim_{x\to\frac{1}{2}}\frac{x-\frac{1}{2}}{\arcsin x-\arcsin\frac{1}{2}}=
\lim_{y\to\frac{\pi}{6}}\frac{\sin y-\frac{1}{2}}{y-\frac{\pi}{6}}
$$
Now set $y-\frac{\pi}{6}=z$ and your limit becomes
$$
\lim_{z\to0}\frac{\sin(z+\frac{\pi}{6})-\frac{1}{2}}{z}
$$
that easily reduces to a couple of fundamental limits by expanding the sine at the numerator.

As usual, I find such kind of exercises a bad way to teach limits.
A: Make the substitution $y=\arcsin x$ and use the sum to product rule to reduce the limit to 
$$
\lim_{t\to 0} \frac{\sin t}{t} = 1.
$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{I \equiv \lim_{x \to 1/2}\left[%
{\arcsin\left(x\right) - \arcsin\left(1/2\right) \over x - 1/2}\right]\,,\qquad
 \mu \equiv \arcsin\pars{1/2}.}$

\begin{align}
I&= \lim_{x \to \mu}\bracks{x - \mu \over \sin\pars{x} - 1/2}
=
\lim_{x \to 0}\bracks{x \over \sin\pars{x + \mu} - 1/2}
\\[3mm]&=
\lim_{x \to 0}
\bracks{x \over \sin\pars{x}\cos\pars{\mu} + \cos\pars{x}\sin\pars{\mu} - 1/2}
=
\lim_{x \to 0}\bracks{x \over \sin\pars{x}\cos\pars{\mu}}
\\[3mm]&=
{1 \over \root{1 - \sin^{2}\pars{\mu}}}\,
\lim_{x \to 0}{1 \over \sin\pars{x}/x} = {1 \over \root{1 - \pars{1/2}^{2}}}
\end{align}

$$\color{#0000ff}{\large%
\lim_{x \to 1/2}\left[%
{\arcsin\left(x\right) - \arcsin\left(1/2\right) \over x - 1/2}\right]=
{2\root{3} \over 3}}
$$
