Interchange of partial derivative and limit

In general: No, you can't. Consider $$ \gamma(T, m) = \frac 1T \sin(T^2 m) $$ Then $\lim_T \gamma(T, m) = 0$, hence $\partial_m \lim_T \gamma(T, m) = 0$. But $\partial_m \gamma(T,m) = T\sin(T^2m)$ and $\lim_T \partial_m \gamma(T,m)$ does not exist.