How to find this limit? I have the limit $$\lim_{x\to0}\frac{\pi/2-\arccos{(x^2)}-x^2}{x^6}.$$ Can somebody please explain to me why this limit exists (according to wolfram alpha it is $1/6$)? Using the standard properties for limits can I not reason that the limit is equal to $$\frac{\pi/2-\lim_{x\to0}\arccos{(x^2)}-\lim_{x\to0}x^2}{\lim_{x\to0}x^6}=\frac{\pi/2-\pi/2-\lim_{x\to0}x^2}{\lim_{x\to0}x^6}=\lim_{x\to0}\frac{-1}{x^4}$$ which is infinity/does not exist. Where exactly is the flaw in my reasoning?
Thanks
 A: (1)Let $\arcsin x^2=y\implies x^2=\sin y$ and $\arccos x^2=\frac\pi2- y$
$$\lim_{x\to 0}\frac{\frac{\pi}2 -\arccos x^2-x^2}{x^6}=\lim_{y\to0}\frac{y-\sin y}{\sin^3y}$$
(2)Alternatively, let $\arccos x^2=y\implies x=\cos y$ and as $x\to0,y=\arccos x^2\to\frac\pi2$
$$\lim_{x\to 0}\frac{\frac{\pi}2 -\arccos x^2-x^2}{x^6}$$
$$=\lim_{y\to \frac\pi2}\frac{\frac\pi2-y-\cos y}{\cos^3y}$$
$$=\lim_{z\to 0}\frac{z-\sin z}{\sin^3z} $$ (Putting  $ z=\frac\pi2-y,y\to\frac\pi2\implies z\to0,\cos y=\cos\left(\frac\pi2-z\right)=\sin z)$ 
Now $$\lim_{y\to 0}\frac{y-\sin y}{\sin^3y}$$
$$=\lim_{y\to 0}\frac{y-(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots)}{y^3}\left(\frac{y}{\sin y}\right)^3$$
$$=\lim_{y\to 0}\frac{(\frac{y^3}{3!}-\frac{y^5}{5!}+\cdots)}{y^3} \text { as  } \lim_{t\to0}\frac {\sin t}t=1$$
$$=\frac16$$
As $$\lim_{z\to 0}\frac{z-\sin z}{\sin^3z} \text {is of the form }\frac00$$ the problem can also be handled using L'Hospital Rule as follows:
$$\lim_{z\to 0}\frac{z-\sin z}{\sin^3z}$$
$$=\lim_{z\to 0}\frac{1-\cos z}{3\sin^2z\cos z} \text {which is again of the form }\frac00$$
$$=\lim_{z\to 0}\frac{\sin z}{-3\sin^3z+3\cos z2\sin z\cos z} \text {which is again of the form }\frac00$$
$$=\lim_{z\to 0}\frac{1}{-3\sin^2z+6\cos^2z}\text{ as } z\to0\implies z\ne0\implies \sin z\ne0 $$
$$=\frac16$$
A: There is a shorter proof is you notice that $\arccos x^2$ can be expanded in Taylor series as $x \to 0: \ \arccos x^2 = \frac{\pi}{2}-x^2-\frac{x^6}{6} + O(x^{10})$. Then the constant and $x^2$ in the numerator cancel out and you remain with 
$$
\lim_{x \to 0} \frac{x^6 + O(x^{10})}{6 x^6}=\frac{1}{6}
$$  
A: First recall that $\dfrac{\pi}2 - \arccos(y) = \arcsin(y)$. Hence, the limit you want to evaluate is
$$L = \lim_{x \to 0} \dfrac{\arcsin(x^2) - x^2}{x^6} = \lim_{t \to 0} \dfrac{t-\sin(t)}{\sin^3(t)} = \lim_{t \to 0} \dfrac{t-\sin(t)}{t^3} \lim_{t \to 0} \left(\dfrac{t}{\sin(t)} \right)^3 = \dfrac16 \times 1 = \dfrac16$$
where $\lim_{t \to 0} \dfrac{t-\sin(t)}{t^3} = \dfrac16$ is proved here.
