I'm working on the F-signature of triples $(R, \Delta, f^t)$ in the case $\Delta = 0$. More precisely, I need to show that for a regular ring R with maximal ideal m, and for any principal ideal f of R, and any $0 \leq t \leq 1$, $$\lim\limits_{e \to \infty} p^{-ed} [l(R/(m^{[p^e]} : f^{\lceil t(p^e-1)\rceil})) - l(R/(m^{[p^e]} : f^{\lceil tp^e\rceil}))] = 0$$ Where $l(\cdots)$ is the length of R-modules, p is a prime number denoting the characteristic of my ring, $d$ is the (Krull) dimension of my ring, and $m^{[p^e]}$ denotes the bracket power of m, i.e. if $m = (x_1, \cdots, x_d)$, then $m^{[p^e]} = (x_1^{[p^e]}, \cdots, x_d^{[p^e]})$. This need to not be mistaken with $\lceil tp^e\rceil$ which is only the entire ceiling value of $tp^e$. Well I have already shown using exact sequences that the limit above is equal to $$\lim\limits_{e \to \infty} p^{-ed} [l(R/(m^{[p^e]} + f^{\lceil tp^e\rceil})) - l(R/(m^{[p^e]} + f^{\lceil t(p^e-1)\rceil}))]$$ which is equal to $$\lim\limits_{e \to \infty} p^{-ed} l(\cfrac{m^{[p^e]} + f^{\lceil t(p^e-1)\rceil}}{m^{[p^e]} + f^{\lceil tp^e\rceil}})$$ if this length is finite. Then, I don't know how to go any further. I tried to show that this length grows as fast as $p^{e(d-1)}$ but I failed. One thing that you might need to help me is that $$l(R/m^{p^e}) = l(R/(x_1^{[p^e]}, \cdots, x_d^{[p^e]})) = p^{ed}l(R/(x_1, \cdots, x_d)) = p^{ed} l(R/m) = p^{ed}$$ because $R/m$ is a field so its length is 1.
Thank you for any help that you can provide !
