Quantum derivatives: quantum calculus and their classical limit

Question

Jackson derivatives and their $q,p$-version are defined to be

$$D_qf=\dfrac{f(qx)-f(x)}{(q-1)x}$$

$$D_{q,p}f=\dfrac{f(qx)-f(px)}{(q-p)x}$$

When trying to go $q\rightarrow 1$ and $q\rightarrow p$ I should get $f'(x)=df/dx$, but applying L'Hôpital rule ($q$ is the variable in the limit) I get

$$D_qf=x\dfrac{df(qx)}{dq}$$

and

$$D_{q,p}f=\dfrac{df(qx)}{dq}x$$

Why am I not getting it right? Is q not a variable? Should I take $x$ instead? In the single variable I consider the alternative, and make a Taylor expansion of f(qx), then I get with q=1:

$$ D_qf=\dfrac{f(qx)+(qx-x)f'(qx)+(qx-x)^2f''(x)-f(x)}{(qx-x)}=Df(x)$$

In some books of quantum calculus I get the expression por $qp$-derivative to be

$$D_{q\rightarrow p}=Df(px)/p=Df(x)=f'(x)$$

Why L'Hôpital fails here and why $d/dq$ is not $d/dx$?

Answer 1

The comment by @htf addresses your misconstrued L'Hôpital's rule, your mishandling of the denominator.

It should be easier for you to define $q=1+\epsilon$ and Taylor expand all around $\epsilon =0$, $$ f(qx) = f(x) + \epsilon x f'(x) + O(\epsilon^2) \implies \\ D_q f(x)= f'(x) + O(\epsilon) \to f'(x),$$ as in WP, which gives the more elegant limit, $$ D_q= \frac{1}{x}~\frac{(1+\epsilon)^{d~~~ \over d (\ln x)} -1}{ \epsilon} = \frac{d}{dx} +O(\epsilon)\to \frac{d}{dx} . $$

Likewise, defining $q=p+\epsilon $, instead, and taking $\epsilon \to 0$ before $p\to 1$, $$D_{q,p}f=\dfrac{f(px) +\epsilon x f'(px) +O(\epsilon^2) -f(px)}{ \epsilon x}= f'(px) + O(\epsilon) \to f'(px) \to f'(x).$$ The second limit is only correct if $p\to 1$ as well. WP deals with the various definitions and limits.