Find $\lim _{x\to \infty }\left(x\left(\arctan2x\:-\arccos\left(\frac{1}{x}\right)\right)\right)$ Find $$\lim _{x\to \infty }\left(x\left(\arctan(2x)-\arccos\left(\frac{1}{x}\right)\right)\right)$$
My idea was to let $t=\frac{1}{x}$ then I get 
$$\lim _{t\to 0 }...$$ 
but i did not know what to do with $\arctan(2x)$
any hint how to solve this ? 
thanks
 A: Using your idea, the limit equals
$$\lim\limits_{x\to 0^+}\frac{\arctan(2/x)-\arccos x}{x}=\lim\limits_{x\to 0^+}\frac{1}{\sqrt{1-x^2}}-\frac{2}{x^2(1+\frac{4}{x^2})}$$ (L'Hoptial's rule)
$$=1-\lim\limits_{x\to 0^+}\frac{2}{x^2+4}=\frac{1}{2}.$$
A: The parantheses goes to $0$, so using $\lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1$, the limit equals:
$$\lim _{x\to \infty }\left(x\left(\arctan 2x-\arccos \frac{1}{x}\right)\right)=\lim_{x\to \infty}x\left(\sin\left(\arctan 2x-\arccos\frac{1}{x}\right)\right)$$
and using the trigonometric identities:
$$\sin(a-b)=\sin a\cos b-\sin b\cos a,\\ \ \arctan x=\arcsin \frac{x}{\sqrt{1+x^2}},\ \arccos x=\arcsin \sqrt{1-x^2}$$
the limit is:
$$
\begin{aligned}
\lim_{x\to \infty}x\sin\left(\arctan 2x -\arccos \frac{1}{x}\right)
&= \lim_{x\to \infty} x\left(\frac{2x}{\sqrt{1+x^2}}\cdot \frac{1}{x}-\frac{1}{\sqrt{1+4x^2}}\cdot \sqrt{1-\frac{1}{x^2}}\right)\\
&=\lim_{x\to \infty} \frac{2-\sqrt{1-\frac{1}{x^2}}}{\sqrt{\frac{1}{x^2}+4}}\\
&=\frac{1}{2}
\end{aligned}
$$
A: Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
$\arctan(2x)-\arccos(1/x)$
$=\pi/2-$arccot$(2x)-(\pi/2-\arcsin(1/x)$
$=\arcsin(1/x)-\arctan(1/2x)$
Now set $1/x=t$ 
Method$\#1$ use $$\lim_{t\to0}\dfrac{\arcsin t}t=1$$ and $$\lim_{t\to0}\dfrac{\arctan(t/2)}{t/2}=1$$
Method$\#2:$
Alternatively apply https://mathworld.wolfram.com/SeriesExpansion.html to find the numerator $$=t-t/2+O(t^3)$$
