On this page
https://en.wikipedia.org/wiki/Rotation_operator_%28quantum_mechanics%29
under "The translation operator," they use Taylor expansion. As part of that proof they state
$p_x = ih * dT(0)/da$
By T(0) they do not mean T evaluated at 0; they mean the translation operator that does nothing.
But what do they MEAN by $p_x = ih * dT(0)/da$? Can't that be read as "the x momentum operator is a scalar multiple of the rate of change in the translation operator, if upgraded by an infinitesimally small amount da, to a 'bigger' translation operator"? What does that mean?
It makes a little more sense if I think about it as:
Our goal is to create an operator that shifts the entire wave function over by an amount (translation). We have an operator that is essentially the derivative (momentum operator).
Let's say we want to shift it over by an amount $\Delta x$. A first approximation would be to create a function that returns the value $\psi(x_0) + (\Delta x * d\psi/dx (x_0))$ at every point $x_0$.
Then this is a very crude approximation of the original $\psi(x)$ shifted over by $\Delta x$.
How could we make the approximation better? Well instead of only using the first derivative once, what if we used the first derivative to get us to $x$ + (some portion of $\Delta x$) say just $1/10 * \Delta x$. Then we have a new $\psi_1$ function which took care of "10%" of the total translation.
Then apply the system again: Take the first derivative of $\psi_1$ again at every point, and then use that to approximately bump every point over by another $1/10 * \Delta x$. That gives us $\psi_2$ which is shifted over now by 20% of the total goal. And what if we did this all the way to $\psi_{10}$.
Well that would make a very good approximation for $\psi$ shifted over by the full $\Delta x$, and the expression representing all that process would be:
$(1 + (1/10 * \Delta x) * d\psi/dx)(1 + (1/10 * \Delta x) * d\psi/dx) .. (1 + (1/10 * \Delta x) * d\psi/dx)$, right? Which if you expand it out with the binomial theorem gets you something that looks very much like the Taylor expansion of e. Hence if you did this to a finer and finer degree, you'd eventually end up with $e^{(\Delta x) d/dx}$. Which is basically what we write as the formula for the translation operator, $e^{\Delta x (i/\hbar)\hat{p} }$.