$f$ weakly modular. How to prove that the order of vanishing $v_p(f)$ is well defined for $p\in SL_2(\mathbb{Z}) \setminus H$ I'm studying the proof of Lemma 1.6.6 in this Lecture notes page 16 


Lemma If $f$  is weakly modular of weight $k,$ the order of vanishing $
v_p(f)$ is well defined for $p\in SL_2(\mathbb{Z}) \setminus
 H.$


I have difficulty to understand the proof, can someone explain to me how to prove this ?
Thanks !
 A: I'm going to replace $z$ by $\gamma z$ on the left side of the original question, so that i can replace the limit $z \to \gamma p$ by $z \to p$. So it becomes 
$\lim_{z\to p}(\gamma z- \gamma p)^{-n}f(\gamma z)=\lim_{z\to p}(\gamma z- \gamma p)^{-n}j(\gamma, p)^kf(z)$
If $\gamma z= \frac{az+b}{cz+d}$, then doing the algebra (making common denominator etc) shows
$\gamma z -\gamma p =\frac{(z-p)(ad-bc)}{(cz+d)(cp+d)}=\frac{(z-p)}{(cz+d)(cp+d)}$
since $ad-bc=1$
Hence the limit becomes 
$$\lim_{z \to p}\frac{(z-p)^{-n}}{(cz+d)^{-n}(cp+d)^{-n}}j(\gamma, p)^kf(z)$$
$$=\lim_{z \to p}(cz+d)^n(cp+d)^nj(\gamma, p)^k(z-p)^{-n}f(z)$$
$$=\lim_{z \to p}(cz+d)^n(cp+d)^nj(\gamma, p)^k(z-p)^{-n}f(z)$$
$$=\lim_{z \to p}j(\gamma, p)^{2n+k}(z-p)^{-n}f(z)$$
because when $z \to p$, $cz+d \to cp+d =j(\gamma, p)$.
Then all this implies the order at $\gamma p$ and order at $p$ are the same because (by definition of order) the order is the largest number $n$ such that each side (of the line in the original question) is non-zero; we've shown the sides are equal for every $n$.
