I am given $ m<<M $. I have to find approximate value of $(\sqrt{\frac{1}{m}+\frac{1}{M}}-\sqrt{\frac{1}{m}})^2$
My book does it as follows:
$ (\sqrt{\frac{1}{m}+\frac{1}{M}}-\sqrt{\frac{1}{m}})^2 = (\sqrt{\frac{1}{m}}\sqrt{1+\frac{m}{M}}-\sqrt{\frac{1}{m}})^2 \tag{1} $
$ = \frac{1}{m}(\sqrt{1+\frac{m}{M}}-1)^2 \tag{2} $
$ =\frac{1}{m}(1+\frac{m}{2M}-1)^2 \tag{3} $
$ =\frac{m}{4M^2} \tag{4} $
I understand this procedure. But i tried to do it differently:
$ (\sqrt{\frac{1}{m}+\frac{1}{M}}-\sqrt{\frac{1}{m}})^2 = (\sqrt{\frac{1}{m}+\frac{1}{M}})^2 + (\sqrt{\frac{1}{m}})^2 - 2\sqrt{\frac{1}{m}+\frac{1}{M}}\sqrt{\frac{1}{m}} \tag{5} $
$ =\frac{1}{m}+\frac{1}{M}+\frac{1}{m}-2\sqrt{\frac{1}{m}}\sqrt{1+\frac{m}{M}}\sqrt{\frac{1}{m}} \tag{6} $
$ =\frac{2}{m}+\frac{1}{M}-2\frac{1}{m}\sqrt{1+\frac{m}{M}} \tag{7} $
$ =\frac{2}{m}+\frac{1}{M}-2\frac{1}{m}(1+\frac{m}{2M}) \tag{8} $
$ =\frac{2}{m}+\frac{1}{M}-\frac{2}{m}-\frac{2m}{2Mm} \tag{9} $
$ =\frac{1}{M}-\frac{1}{M} = 0 \tag{10} $
I am getting 0. Where did i make an error.