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In this post, Fredrik Meyer gives a proof to the following formula(Please see the conditions and the meaning of notations in the link):

$\frac{d}{dt}|_{t=0} \alpha(t) = [X,Y](p)$, where $\alpha(t):= \phi_{-\sqrt{t}}^Y \circ \phi_{-\sqrt{t}}^X \circ \phi_{\sqrt{t}}^Y \circ \phi_{\sqrt {t}}^X$.

I think Fredrik's proof is nice everywhere except for here(the last formula)

Now we use that $(\phi_h^X)^{-1}=\phi_{-h}^X$ to get from $ \lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) - f \circ \phi_h^X \circ \phi_h^Y(p) \right] $ \begin{equation} \lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right] \tag{1} \end{equation}

I wonder why this is correct as given. I think by using that $(\phi_h^X)^{-1}=\phi_{-h}^X$ we can get

$$\lim_{h \to 0} \frac {1}{h^2}\left[ f\circ \phi_h^Y \circ \phi_h^X(p)- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y \circ \phi_h^Y \circ \phi_h^X(p) \right] \tag{2}$$

But it is hard to see why (2) actually implies (1). So I even suspect that his proof is not correct.Can anybody explain it to me or show me how to correct it?

Intuitively $\phi_h^Y \circ \phi_h^X(p)$ somehow "converges" to $p$. But in the formal proof, I am afraid we can't take the limit of this part first.

Ps: I haven't read through this proof given by Owen Barrett, but I feel like the proof using Taylor expansion is too long and prefer a direct proof.

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