Determining the behavior of the transference function as $\omega\to 0$ or $\omega\to\infty$ I'm studying experimental Physics and one of the experiments I must understand is about electronics concerning filters. Considering one RLC filter, measuring the output at the resistor we have the transference function
$$H(\omega)=\dfrac{R}{R+j(X_L-X_C)}$$
Being $X_L=\omega L$ the reactance of the inductor and $X_C=1/\omega C $ the reactance of the capacitor. The book considers then the limit cases $\omega\to 0$, $\omega = \omega_0$ and $\omega\to \infty$ being $\omega_0 = 1/\sqrt{LC}$. The case $\omega = \omega_0$ is pretty easy. In that case, $H(\omega_0) = 1$ and I'm fine with that.
The problem is that the book says that if $\omega \to 0$ we have
$$H(\omega)=j\omega RC$$
And if $\omega\to \infty$
$$H(\omega) = -\dfrac{jR}{\omega L}$$
This is pretty confusing and I don't really know how I can get to that. The book says all of this comes from the fact that if $\omega \to 0$ then $X_C >> X_L$ and if $\omega \to \infty$ we have $X_L >> X_C$. This is obvious from the definitions of the reactances, but how this gives these results?
Also, I'm pretty sure this really aren't limits. When we consider $\lim_{x\to a}f(x)$ the limit of some function $f$, we don't expect the answer to depend on $x$, however on these cases the answer do depend on $\omega$.
So, what is really going here and how can we derive those results?
 A: You're right that the expressions for $H(\omega)$ as $\omega\to 0$ and $\omega\to\infty$ are not really limits, at least not as mathematicians use the word. They're leading-order series expansions, in the parameter $\omega$ (for $\omega\to 0$) or $\frac{1}{\omega}$ (for $\omega\to \infty$). You can think of these expansions as the "limiting behavior" of $H(\omega)$ as it approaches either of the points $0$ or $\infty$, which is why physicists often use the word "limit" in this context.
There are two ways to derive the results. The brute-force way (which is really not very complicated, and you should work it out yourself) is to just plug the definitions of $X_L$ and $X_C$ into $H(\omega)$ and compute the first term of the relevant series. For $\omega\to\infty$ it may help to first rewrite $H$ in terms of $u = \frac{1}{\omega}$.
The other way, the "shortcut", is to conclude that if $X_L$ (for example) is very large, it doesn't much matter what the values of $R$ or $X_C$ are, and thus
$$R + j(X_L - X_C) \to jX_L$$
Then just simplify the resulting fraction. And then you can do the same for the case where $X_C$ is the large variable.
A: The transfer function given is for a bandpass filter.  Multiply the numerator and denominator by $j\omega C$ to yield a more standard form for a bandpass filter:
$$H(j \omega) = \frac{j\omega RC}{(1 - \omega^2LC) + j\omega RC}$$
Now, it's easy to see that, for $\omega$ 'small enough' (such that the $1$ in the denominator is much larger than the frequency dependent terms),
$$H(j \omega) \approx j\omega RC$$
which goes to zero as $\omega \rightarrow 0$ as desired for a bandpass filter.
Similarly, for $\omega$ 'large enough' (such that the squared frequency dependent term in the denominator is much larger than the other terms),
$$H(j \omega) \approx \frac{j\omega RC}{-\omega^2LC} = -\frac{jR}{\omega L}$$
which goes to zero as $\omega \rightarrow \infty$ as desired for a bandpass filter.
Put another way, at the frequency extremes, the transfer function asymptotically approaches the low and high frequency approximations given.
