How can we interpret $\lim_{\beta l \to \pi/2} \tan (\beta l)$ in the quarter-wave impedance transformer equation?

Question

From the theory of Quarter wave impedance transformer wikipedia article, it is written $$ \begin{align*} Z_{in} &=\lim_{\beta l \to \pi/2}Z_0\frac{Z_l + j Z_0 \tan (\beta l)}{Z_0 + j Z_l \tan (\beta l)}\\ &=Z_0\frac{j Z_0 }{ j Z_l} \end{align*} $$ Does that mean that $$ \lim_{\beta l \to \pi/2} \tan (\beta l) = +\infty $$ and hence $$ \begin{align*} Z_{in} &=\lim_{\beta l \to \pi/2}Z_0\frac{\underbrace{Z_l}_{\text{relatively small}} + j Z_0 \underbrace{\tan (\beta l)}_{\text{relatively big}}}{\underbrace{Z_0}_{\text{relatively small}} + j Z_l \underbrace{\tan (\beta l)}_{\text{relatively big}}}\\ \implies Z_{in} &=Z_0\frac{j Z_0 }{ j Z_l} \end{align*} $$ ?

Answer 1

The interpretation is that by adding a quarter wavelength of transmission line to the original load impedance, $Z_l$, it is transformed to its dual impedance $Z_0^2/Z_l$. So, for example, a purely capacitive impedance is transformed into a purely inductive one, and vice versa.