Reason: (√( 1-cosx))/x

=(√(2(sin(x/2))^2)/x

=(√2)|sin(x/2)|/x

X→0-

(√( 1-cosx))/x =(-√2/2)(sin(x/2)/(x/2)→-√2/2

When x→0+

(√( 1-cosx))/x =(√2/2)(sin(x/2)/(x/2)→√2/2

That is, f(x) exists at the left and right poles of x=0, but it is not equal, so it is the first kind of insoluble (jumping) discontinuity.

I hope I can help you!