What does the limit $\lim_{x \to a}\frac{(x-a)f'(x)}{f(x)}$ mean for non-polynomials The limit $$\lim_{x \to a}\frac{(x-a)f'(x)}{f(x)}$$
has an important meaning for polynomials. Suppose $f(x)=(x-a)^nq(x)$ where $q(a)\neq 0$. Then, the given limit is equivalent to
$$\lim_{x \to a}\frac{\{n(x-a)^{n-1}q(x)+(x-a)^nq'(x)\}(x-a)}{(x-a)^nq(x)}=\lim_{x \to a}\frac{nq(x)+(x-a)q'(x)}{q(x)}=n+\lim_{x \to a}\frac{(x-a)q'(x)}{q(x)}=n$$
where the last limit comes from the fact that $q(a)\neq 0$, but the numerator goes to zero. So, for polynomials, the value of the given limit would indicate the algebraic multiplicity of the polynomial of the term $(x-a)$. We could also extend this concept to functions not quite polynomial, but "looks" like a polynomial, for example functions like $x-4\sqrt{x}+3$. This function has the above limit value when $a=1$ as
$$\lim_{x \to 1}\frac{(x-1)\times \{1-\frac{2}{\sqrt{x}}\}}{(\sqrt{x}-1)(\sqrt{x}-3)}=\frac{2\cdot (-1)}{1-3}=1$$
One could also extend to functions like $f(x)=x^\pi$. The limit in this case would be,
$$\lim_{x \to 0}\frac{x\times \pi x^{\pi-1}}{x^\pi}=\pi$$
So, the limit would similarly be the value of the "degree of $(x-a)$"(which, if it is even a valid saying).
(Note that we choose function $f(x)$ and the value $a$ for each limit. Yet since the limit has some meaning when $f(a)=0$, I would omit the choice of $a$ whenever the root of $f(x)$ is unique.)
My question is, how can I extend this concept to functions that are not in the form of $f(x^p)$(where $f$ is a polynomial, $p$ is a real number)? Are there any meaning to this limit value when the function is a transcendental function? Are there any concepts that this value represents?

(Edit) Here are some examples of the value of the limit for some functions.

*

*$f(x)=e^{-x}(x^3\sin x)$ at $a=0$ $$\lim_{x \to 0}\frac{xf'(x)}{f(x)}=\lim_{x \to 0}\frac{x\times e^{-x}(-x^3\sin x+3x^2\sin x+x^3\cos x)}{e^{-x}(x^3\sin x)}=4$$

*$f(x)=\frac{\sin x}{x}$ at $a=0$ $$\lim_{x \to 0}\frac{xf'(x)}{f(x)}=\lim_{x \to 0}\frac{x\times \frac{x\cos x-\sin x}{x^2}}{\frac{\sin x}{x}}=0$$

*$f(x)=\frac{\sin x}{x}$ at $a=\pi$ $$\lim_{x \to \pi}\frac{(x-\pi)f'(x)}{f(x)}=\lim_{x \to \pi}\frac{(x-\pi)\frac{x \cos x-\sin x}{x^2}}{\frac{\sin x}{x}}=1$$

*$f(x)=e^x-1$ at $a=0$ $$\lim_{x \to 0}\frac{xf'(x)}{f(x)}=\lim_{x \to 0}\frac{x\times e^x}{e^x-1}=1$$

If $\ f\ $ is at least $\ n\ $ times continuously differentiable in a neighbourhood of $\ a\ $, $\ f^{(n)}(a)\ne0$$\,\ $, and $\ f^{(k)}(a)=0\ $ for all $\ k\le n-1\ $, then by Taylor's theorem with Lagrange form of remainder,
\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{(x-a)^if^{(i)}(a)}{i!}+\frac{(x-a)^nf^{(n)}(\xi)}{n!}\\
&=\frac{(x-a)^nf^{(n)}(\xi)}{n!}\\
f'(x)&=\sum_{i=0}^{n-2}\frac{(x-a)^if^{(i+1)}(a)}{i!}+\frac{(x-a)^nf^{(n)}(\eta)}{(n-1)!}\\
&=\frac{(x-a)^nf^{(n)}(\eta)}{(n-1)!}\ ,
\end{align}
where $\ a\le\xi,\eta\le x\ $ or $\ x\le\xi,\eta\le a\ $.  It follows that for any such $\ f\ $
$$
\lim_{x\rightarrow a}\frac{(x-a)f'(x)}{f(x)}=n\ .
$$
