Since the ratio of DF to df in the limit is crucial, more detailed analysis is required. From the similar right triangles, D F P F = E D E S and E D 2 = ( D F + F S ) 2 − − E S 2 , giving ( P F 2 − − E S 2 ) D F 2 P F 2 + 2 ⋅ ⋅ F S ⋅ ⋅ D F + F S 2 − − E S 2 = 0 . Solving the quadratic for DF, in the limit as ES approaches FS, the smaller root, D F = E S − − F S . More simply, as DF approaches zero, in the limit the D F 2 term can be ignored: 2 ⋅ ⋅ F S ⋅ ⋅ D F + F S 2 − − E S 2 = 0 leading to the same result. Clearly df has the same limit, justifying Newton’s claim.