I'm currently reading a book on communications circuits and I'm completely stuck on one particular derivation given in the book. This isn't homework and I'm definitely no mathematician. I believe it's more of a maths issue which is why I've posted here as opposed to the electronics stack site.
Here's the section of interest from the book:
Basically, the author claims that for high values of omega, the expression:
$$\Omega = Q\frac{\left ( 1-\frac{1}{\Omega } \right )^{2}}{1-\frac{1} {n\Omega }}$$
reduces to
$$\Omega = Q + \frac{1}{n}$$
and for the life of me, I can't see how! Can anyone shed any light?
$$ \eqalign{ & \Omega = Q{{\left( {1 - {1 \over \Omega }} \right)^{\,2} } \over {\left( {1 - {1 \over {n\Omega }}} \right)}} \cr & {Q \over \Omega }\left( {1 - {1 \over \Omega }} \right)^{\,2} = 1 - {1 \over {n\Omega }} \cr & 1 - {1 \over n}\left( {{1 \over \Omega }} \right) = Q\left( {{1 \over \Omega }} \right)\left( {1 - 2\left( {{1 \over \Omega }} \right) + \left( {{1 \over \Omega }} \right)^{\,2} } \right) \cr & 1 - \left( {{1 \over n} + Q} \right)\left( {{1 \over \Omega }} \right) \approx 0 + O\left( {\left( {{1 \over \Omega }} \right)^{\,2} } \right)\quad \left| {\;{1 \over \Omega } \to 0} \right. \cr & {1 \over \Omega } \approx {1 \over {{1 \over n} + Q}}\quad \left| {\;{1 \over \Omega } \to 0} \right. \cr & \Omega \approx {1 \over n} + Q\quad \left| {\;\Omega \to \infty } \right. \cr} $$