Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\frac{\sin(x)}{x}}$ (I have no clue, what to do next with it), but with Taylor and ended up with:
$$\lim_{x\to 0}\frac{\sqrt{1-\frac{x^2}{6}+o(x^2)}+o(\sqrt{1-\frac{x^2}{6}+o(x^2)})}{x^4} $$
which tends to infinity
 A: \begin{align}
\lim_{x\to 0}\frac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{x^4}
&=\lim_{x\to0}\underbrace{\dfrac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{\left(1-\sqrt{\dfrac{\sin x}{x}}\right)^2}}_{=\frac12}\times\left(\dfrac{1-\sqrt{\dfrac{\sin x}{x}}}{x^2}\right)^2\\
&=\dfrac12\times\left(\lim_{x\to0}\dfrac{1-\sqrt{\dfrac{\sin x}{x}}}{x^2}\right)^2\\
&=\dfrac12\times\left(\lim_{x\to0}\dfrac{x-\sin x}{x^3}\times\dfrac{1}{1+\sqrt{\dfrac{\sin x}{x}}}\right)^2\\
&=\dfrac12\times\left(\dfrac16\times\dfrac12\right)^2\\
&=\boxed{\dfrac1{288}}
\end{align}
A: Hint
Compose Taylor series one piece at the time
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\sqrt{\frac{\sin (x)}{x}}=1-\frac{x^2}{12}+\frac{x^4}{1440}+O\left(x^6\right)$$
$$\cos \left(1-\sqrt{\frac{\sin (x)}{x}}\right)=1-\frac{x^4}{288}+\frac{x^6}{17280}+O\left(x^8\right)$$
A: Since there is $x^4$ on denominator we have to go at least the same order on numerator.
$\sin(x)=x-\frac 16 x^3+\frac 1{120}x^5+o(x^5)$
$\dfrac{\sin(x)}x=1-\frac 16 x^2+\frac 1{120}x^4+o(x^4)$
$S=\left(\frac{\sin(x)}x\right)^\frac 12=1+\frac 12\left(-\frac 16 x^2+\frac 1{120}x^4+o(x^4)\right)-\frac 18\left(-\frac 16 x^2+\frac 1{120}x^4+o(x^4)\right)^2+o(x^4)=1+(-\frac 1{12})x^2+(\frac 1{2\times 120}-\frac 1{8\times 6^2})x^4+o(x^4)$
$\cos(1-S)=\cos(\frac 1{12}x^2-\frac 1{1440}x^4+o(x^4))=1-\frac 12\left(\frac 1{12}x^2-\frac 1{1440}x^4+o(x^4)\right)^2+o(x^4)=1+(\frac {-1}{2\times 12^2})x^4+o(x^4)$
$\dfrac{1-\cos(1-S)}{x^4}=\dfrac{\frac 1{288}x^4+o(x^4)}{x^4}=\frac 1{288}+o(1)\to\frac 1{288}$
Note that many terms disappear quickly because they are negligible compared to $x^4$. In particular only the $-\frac 16x^2$ term of $\sin$ development brings something to the final result, but we need nevertheless to make the calculation up to $o(x^4)$ along the entire chain to ensure coherence.
A: You can reason as follows:
$$\frac{\sin x}x\sim 1-\frac{x^2}{3!}$$ and after taking one minus the square root,
$$\frac{x^2}{12}.$$
Then one minus the cosine yields
$$\frac12\left(\frac{x^2}{12}\right)^2=\frac{x^4}{288}.$$
