In the standard simplified derivation of group velocity (which can be found here) we use two waves $$y_1=A\sin(K_1x-\omega_1 t)$$ $$y_2=A\sin(K_2x-\omega_2 t)$$ In the proof we then get $$V_g=\frac{\Delta \omega}{\Delta k}$$ But I do not understand the step where this is then turned into $$V_g=\frac{d \omega}{d k}$$ why do we assume that $\Delta \omega$ and $\Delta k$ are small? The derivation is valid in the case where they are not small, which means that $$V_g= \frac{d \omega}{d k}$$ does not hold in this case and therefore does not hold in general.
Consider this example
Let $K_1=3$ and $k_2=1$ and let us say we have relationship $\omega=k^3$ using my first fromula we get $V_g=13$ but using the second (with $\bar k=2$) we get $V_g=12$, theses are diferent.
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