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

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.
