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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 ?

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  • $\begingroup$ The question is not clear. Consider using Latex or at least putting Brackets $\endgroup$ – LeeNeverGup Nov 23 '13 at 20:57
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    $\begingroup$ This is finding the derivative of $\arcsin$ at $1/2$. $\endgroup$ – egreg Nov 23 '13 at 21:06
  • $\begingroup$ @Lucian That was not an answer: using the derivative is just using a special case of l'Hôpital's theorem, isn't it? $\endgroup$ – egreg Nov 23 '13 at 21:37
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So that there won't be complains:

$$\lim_{x\to 1/2}\frac{\arcsin x-\arcsin\frac12}{x-\frac12}=:\left(\arcsin\frac12\right)'=\frac1{\sqrt{1-\frac12}}$$

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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 $$\lim_{x\to\frac{1}{2}}\frac{\arcsin x-\arcsin\frac{1}{2}}{x-\frac{1}{2}}=\arcsin'(\frac12)=\frac1{\sqrt{1-(\frac12)^2}}=\frac1{\sqrt{\frac34}}=\frac2{\sqrt3}$$

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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.

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Make the substitution $y=\arcsin x$ and use the sum to product rule to reduce the limit to $$ \lim_{t\to 0} \frac{\sin t}{t} = 1. $$

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  • $\begingroup$ Until I googled it some 2 minutes ago I never knew that Prosta...whatever (simple trigonometric identities), but the limit is not $\;1\;$ ... $\endgroup$ – DonAntonio Nov 23 '13 at 21:20
  • $\begingroup$ I added the correct reference. I realized that "prostaferesi" was the italian name... The result is not 1 but you reduce your computation to that one. $\endgroup$ – Emanuele Paolini Nov 23 '13 at 21:25
  • $\begingroup$ It still is confusing and misleading, imo: what you get is the limit $$\lim_{y\to \pi/6}\frac{y-\pi/6}{\sin y-1/2}$$ and this limit is not one... $\endgroup$ – DonAntonio Nov 23 '13 at 21:29
  • $\begingroup$ Transform $\sin y - \sin\pi/6$ into a product... $\endgroup$ – Emanuele Paolini Nov 23 '13 at 21:42
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{I \equiv \lim_{x \to 1/2}\left[% {\arcsin\left(x\right) - \arcsin\left(1/2\right) \over x - 1/2}\right]\,,\qquad \mu \equiv \arcsin\pars{1/2}.}$

\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}

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

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