This is a doubt I got in the following question: Set of all values of x so $$ \lim_{x \to \infty} \left(\frac{1}{1+\frac{4\tan^{-1} (2x)}{\pi}}\right)^2n $$ is non zero and finite, n is a natural number . Answer is [-1/2,1/2].
The answer says $4\tan^{-1} (2x)/\pi$ has range -1 to 1 . How is this possible when range of $\tan^{-1} x$ is $-\pi/2$ to $\pi/2$?