Why does $y = \frac{2A\sin(x\pi \ell)}{\pi^2 \ell (1-\ell) x^2}$ simplify to $y=c/x$ as $\ell$ approaches $0$?

Question

I am working on something using this equation, and I find something strange. I am manipulating $\ell$ here between $0$ and $1$. I note that as $\ell$ approaches zero ($\sim0.001$ or less) it becomes a graph of $y=c\frac{1}{x}$. I cannot understand why, or how $c$ would be calculated as a simplification/approximation of the equivalent full formula at this point.

The full equation is: $$y = \frac{2A\sin(x\pi \ell)}{\pi^2 \ell (1-\ell) x^2}$$

As you can see as $\ell$ is close to $0$, this becomes equivalent to $y=c\frac{1}{x}$:

However, I don't know how this is happening, or what $c$ value would then reliably approximate the full equation at a certain value of $A$ (I just matched the curves visually here with an arbitrary constant).

Any help understanding the simplification and thus how $c$ could be calculated in terms of $A$ to roughly match in these circumstances?

Answer 1

Recall $\sin t \approx t$ for $t$ near $0$, so your equation can be approximated by $$y = \frac{2A\sin(x\pi l)}{\pi^2 l (1-l) x^2} $$ $$\approx \frac{2Ax\pi l}{\pi^2 l (1-l) x^2}$$ $$=\frac{2A}{\pi (1-l) x}$$ $$\approx \frac{2A}{\pi}\cdot \frac1x$$ (this last approximation because $1-l\approx 1$)

Answer 2

When $x$ is close to $0$, $$\frac{\sin x}x\approx 1$$ In your case, use $$\frac{\sin x\pi l}{x\pi l}=1$$ Therefore $$y=\frac{2A\sin(x\pi l)}{\pi^2 l(1-l)x^2}=\frac{2A}{\pi(1-l)x}\frac{\sin x\pi l}{x\pi l}\approx\frac{2A}{\pi(1-l)}\frac 1x$$ Note that for large $x$ values, or larger $l$, the function will oscillate around $0$.

Answer 3

For fixed $x \ne 0$,

$$\begin{align*} \lim_{l\to0}\left(\frac{2A}{\pi x}\cdot \frac1{1-l}\cdot\frac{\sin(x\pi l)}{\pi l x}\right) &= \lim_{l\to0}\left(\frac{2A}{\pi x}\right) \cdot \lim_{l\to0}\left(\frac1{1-l}\right) \cdot \lim_{l\to0}\left(\frac{\sin(x\pi l)}{\pi l x}\right)\\ &= \frac{2A}{\pi x}\cdot 1 \cdot 1 \end{align*}$$