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I have since seen more concise proofs, but am curious if this works. If f is differentiable at an interior point $z_0$, then: $\forall\epsilon>0,\exists \delta_1>0$ such that

$|z-z_0|<\delta_1\implies|\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)|<\sqrt{\epsilon/2}$.

My main concern is this part; is it valid to have $\sqrt{\epsilon/2}$ rather than $\epsilon$. I've seen $c\epsilon$ used with $c>0$. I thought doing this was valid because $\epsilon$ and $c\epsilon$ have the same range for $\epsilon>0$. If this is the case $\sqrt{\epsilon/2}$ should be valid too right? I also use $\lim_{z\to z_0}(z-z_0)=0$:

$\forall\epsilon>0,\exists \delta_2>0$ such that $|z-z_0|<\delta_2\implies|z-z_0|<\sqrt{\epsilon/2}$

and $\lim_{z\to z_0}f'(z_0)(z-z_0)=0$:

$\forall\epsilon>0,\exists \delta_3>0$ such that $|z-z_0|<\delta_3\implies|z-z_0||f'(z_0)|<\epsilon/2$

Let $\delta=min\{\delta_1,\delta_2,\delta_3\}$.

$|f(z)-f(z_0)|=|z-z_0||\frac{f(z)-f(z_0)}{z-z_0}+f'(z_0)-f'(z_0)|$

$\leq |z-z_0||\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)|+|z-z_0||f'(z_0)|$

$<\sqrt{\epsilon/2}\sqrt{\epsilon/2}+\epsilon/2=\epsilon$

Thanks

EDIT:

Forgot the final line:

Therefore $\forall \epsilon>0$ $\exists \delta=min\{\delta_1,\delta_2,\delta_3\}$, such that:

$0<|z-z_0|<\delta\implies|f(z)-f(z_0)|<\epsilon$

Therefore differentiability at z_0 implies continuity at z_0

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You proof seems fine.

You might want to be careful with $\color{blue}{0<}|z-z_0|<\delta$.

The reason why you can choose $\sqrt{\frac{\epsilon}2}$ is because it is positive.

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