Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3) I'm having difficulty in finding the following limit.
$$\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
I tried manipulating the given limit in standard limit(s) but I got nowhere. I tried L'Hôpital's rule and then realized it would be very lengthy and daunting so I left it. The answer is 1/2.
Thanks for any help in advance.
 A: for $|x|<1$, $\arcsin(x)=x+x^3/6+3x^5/40+\cdots$, while $\arctan(x)=x-x^3/3+x^5/5+\cdots$, so $(\arcsin(x)-\arctan(x))/x^3=(1/3+1/6)+O(x^2)\to 1/2 $ as $x\to 0$
A: It's worth mentioning that l'Hopital does work here. Your limit becomes
$$\lim_{x \rightarrow 0} {{1 \over \sqrt{1 - x^2}} - {1 \over 1 + x^2} \over 3x^2}$$
$$= \lim_{x \rightarrow 0}{(1 + x^2) - \sqrt{1 - x^2} \over 3x^2(1 + x^2)\sqrt{1 - x^2}}$$
Multiply numerator and denominator by $(1 + x^2) + \sqrt{1 - x^2}$ and you get
$$= \lim_{x \rightarrow 0}{(1 + x^2)^2 - (1 - x^2) \over 3x^2 ((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
$$= \lim_{x \rightarrow 0}{3  + x^2 \over 3((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
You can now just plug in $x = 0$ into the above and get ${1 \over 2}$.
A: Without using L'Hospital's and Series Expansion,
Using this and Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$, 
$$\arcsin x-\arctan x=\arcsin x-\arcsin\frac x{\sqrt{1+x^2}}=\arcsin\left(x\cdot\frac1{\sqrt{1+x^2}}-\sqrt{1-x^2}\cdot\frac x{\sqrt{1+x^2}} \right)$$
$$=\arcsin\left(x\cdot\frac{(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}\right)$$
$$=\arcsin\left(x\cdot\frac{1-(1-x^2)}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\arcsin\left(\frac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)$$
$$\implies\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
$$=\lim_{x\to 0}\frac{\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)}{\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}}\cdot\frac1{\lim_{x\to 0}\sqrt{1+x^2}(1+\sqrt{1-x^2})}$$
Set $\displaystyle\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\theta\implies\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}=\sin\theta$ in the first limit to find the limit to be $1$
The second limit is easy enough
A: If you use L'Hopital's, you must use it multiple times (And the derivatives are quite annoying -- power series is probably the better approach here)

Differentiate the first time
$$\lim \limits_{x \to 0} \dfrac{1/\sqrt{1 - x^2} - 1/(x^2 + 1)}{3x^2} \qquad \text{still is 0/0}$$
$$\lim \limits_{x \to 0} \dfrac{x/(1 - x^2)^{ 3/2} + 2x / (x^2 + 1)^2}{6x} \qquad \text{still is 0/0}$$
$$\lim \limits_{x \to 0} \dfrac{x^2(3/(1 - x^2)^{5/2} - 8/(x^2 + 1)^3) + 1/(1 - x^2)^{3/2} + 2(x^2+1)^2}{6} = \frac{0 + \frac 11 + \frac 21}{6} = \frac{1}{2}$$
