I have a function: $$\rvert H_a(\Omega)\lvert = \frac{1}{\sqrt{1+(\frac{\Omega}{\Omega_c})^{2N}}}$$
My professor says in the function as $\Omega \rightarrow \infty$, $1+(\frac{\Omega}{\Omega_c})^{2N} \rightarrow (\frac{\Omega}{\Omega_c})^{2N}$ and hence $\rvert H_a(\Omega)\lvert \rightarrow (\frac{\Omega}{\Omega_c})^{-N}$ as $\Omega \rightarrow \infty$. which does not make any sense.
As per my thinking, as $\Omega$ tends to $\infty$, $1+(\frac{\Omega}{\Omega_c})^{2N}$ tends to $\infty$ and hence $\rvert H_a(\Omega)\lvert \rightarrow 0$.
Which is the correct result?and if my professor is right can anyone please help me understand the rationale behind it?
The function$\rvert H_a(\Omega)\lvert$ is the magnitude response of a butterworth filter.
First, "$|H_a(\Omega)|\to (\Omega/\Omega_c)^{−N}$" is not a limit statement, since $\Omega$ still appears. Rather, this describes the asymptotic behavior of the function.
As for why $|H_a(\Omega)|\to (\Omega/\Omega_c)^{−N}$ holds ... Large $\Omega$ eventually outpaces any constant $\Omega_c$, so that $\Omega/\Omega_c$ is itself large. Raising a large number to the $2N$ power makes a LARGE number. Adding $1$ to such a thing has virtually-no effect, so you can reasonably ignore it. Doing so, the expression simplifies: $$\frac{1}{\sqrt{1+(\Omega/\Omega_c)^{2N}}}\approx \frac{1}{\sqrt{0+(\Omega/\Omega_c)^{2N}}}=\frac{1}{(\Omega/\Omega_c)^{N}}=(\Omega/\Omega_c)^{−N}$$