The key factor of this function segmentation is the limit of 2n power of x, which is obviously related to the value of base X.

When 0 ≤| x | < 1, the 2n power of x tends to 0, so: f (x) =1+x.

When x=- 1, f (x) = 0/2 = 0;

When x= 1, f(x)=2/2= 1.

When |x| > is at 1, the 2n power of x tends to infinity, so: f(x)=0.

Get segmentation expression:

f(x)= 1

① 0，x∈(-∞，- 1]

② 1+x，x∈(- 1， 1)

③ 1，x= 1

④ 0，x∈( 1，+∞)

Obviously, x= 1 is the discontinuity of the function.