Well, I guess we would tan(arccosx) into sines and cosines: $\frac{ x^{2}\sin(arccosx) }{ \cos(arccosx) }$cos(arccosx) just cancels to x. And then that resulting x cancels out with one of the x's on top to leave: $xsin(arccosx)$Now that we have this, theres nothing undefined to even worry about. Plug in 0 and the limit gives you 0