Evaluating $\lim_{x\to 0^+} \frac1{x} \cos^{-1} (\frac{\sin x}{x})$

Question

Today I came across the following limit problem: $$\lim_{x\to 0^+} \frac1{x} \cos^{-1} \Big(\frac{\sin x}{x}\Big)$$

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Here's my work:

$$\begin{align}L &= \lim_{x\to 0^+} \frac1{x} \cos^{-1}\left (\frac{\sin x}{x}\right) \tag{1} \\& = \lim_{h\to 0}\frac1h \cos^{-1}\left(\frac{\sin h}{h}\right)\tag{2} \\& = \lim_{h\to 0}\frac1h \sin^{-1}\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right) \\ & = \lim_{h\to 0}\frac1h \frac{\sin^{-1}\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right)}{\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right)}\cdot \left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right) \\& = \lim_{h\to 0}\frac{\sin^{-1}\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right)}{\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right)}\cdot \lim_{h\to 0}\frac1h\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right)\tag{3} \\& = \lim_{h\to 0}\frac1h\left(\sqrt{1 - \frac{\sin^2 h}{h^2}}\right) \\& = \lim_{h\to 0} \sqrt\frac{h + \sin h}{h} \cdot \sqrt{\frac{h -\sin h}{h^3}} \\& = \lim_{h\to 0}\sqrt{1 + \frac{\sin h}{h} } \cdot \lim_{h\to 0}\sqrt{\frac{h -\sin h}{h^3}}\tag{4} \\& = \sqrt{2} \cdot \frac1{\sqrt{6}}\\& = \frac1{\sqrt{3}}\end{align}$$

Details:

$(1)$ Putting $x = 0 + h$ as $x \to 0^+$, $h\to 0$.

$(2)$ Used $\cos^{-1}(x) = \sin^{-1}\sqrt{1-x^2}$.

$(3)$ Used $\lim_{x\to 0} \frac{\sin^{-1} x}{x} = 1$.

$(4)$ Used $\lim_{x\to 0}\frac{\sin x}{x}= 1$ for the first limit. And the second limit is evaluated using Maclaurin Series of $\sin h$.

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Although I'm able to solve it but my method is too long. I'm looking for a shorter or alternative way to do it.

Answer 1

Your method is correct and the working can be simplified a lot by writing $(\sin x) /x=\cos t$ so that $t\to 0^+$.

Then we have $$\frac{t} {x} =\sqrt{\frac{t^2}{x^2}}=\sqrt{\frac{t^2}{1-\cos t} \cdot\frac{1-\cos t} {x^2}}$$ The first factor inside the square root tends to $2$ and hence desired limit equals the limit of $$\sqrt{2}\sqrt{\frac{x-\sin x} {x^3}}$$ Thus the desired limit is $\sqrt{2}\sqrt{1/6}=1/\sqrt{3}$.

The square roots are introduced in the working because the most familiar limit associated with $\cos $ function is $$\lim_{t\to 0}\frac{1-\cos t} {t^2}=\frac{1}{2}$$

Answer 2

First note that $\left|\frac{\sin x}x\right|\le1,$ which allows to define $t:=\cos^{-1}\left(\frac{\sin x}x\right).$

As $x\to0,$ we have $t\to0^+,$ $\frac{\sin x-x}{x^3}\to-\frac16$ and $\frac{\cos t-1}{t^2}\to-\frac12,$ hence (using the notion of asymptotic equivalence): $$-\frac{x^2}6\sim\frac{\sin x}x-1=\cos t-1\sim-\frac{t^2}2.$$ Therefore, $$\lim_{x\to0^+}\frac tx=\lim_{x\to0}\sqrt{\frac {t^2}{x^2}}=\sqrt{\frac26}=\frac1{\sqrt3}.$$