Let FX = {[1 + SiNx + (SiNx) ^ 2] ^ 0.5 - (a + bsinx)} / (SiNx) ^ 2, and point x = 0 is the removable discontinuity point of F (x), find a, B The process should be more detailed

Let FX = {[1 + SiNx + (SiNx) ^ 2] ^ 0.5 - (a + bsinx)} / (SiNx) ^ 2, and point x = 0 is the removable discontinuity point of F (x), find a, B The process should be more detailed

Because it is a removable breakpoint, it shows that when x is close to 0, the limit of F (x) exists. When x = 0, the denominator is 0, so the molecule is also 0, and a = 1 is obtained. When x = 0, the denominator is 2sinx * cosx. When x = 0, the value is still 0, which means that the molecule is still 0 when x = 0. When x = 0, the molecular derivation can get b = 1,2