I am given an implicit expression for an algebraic function, ##w(z)## as:##f(z,w)=(1-3 z+3 z^2-z^3)+(-4+8 z-4 z^2)w+(6-6 z)w^2+(-4)w^3+(1)w^4=0## with ##\displaystyle \frac{dw}{dz}=-\frac{\frac{df}{dz}}{\frac{df}{dw}}=\frac{6 w^2+8 w z-8 w+3 z^2-6 z+3}{4 w^3-12 w^2-12 w z+12 w-4 z^2+8 z-4}##. And note at the point ##(1,0)## we have: ##\frac{dw}{dz}=\frac{0}{0}##. However, I know ##w(z)=1+z^{1/4}+z^{1/2}+z^{3/4}## and so obviously the point ##(1,0)## is a removable singularity. In fact, ##\frac{dw}{dz}\biggr|_{(z=1,w=0)}=\{-1/2,-1/2+1/2 i, -1/2-1/2i,3/2\}##(adsbygoogle = window.adsbygoogle || []).push({});

But I only know this because I have the (fractional) polynomial solution and can compute the derivatives directly.

If I didn't know the solution but knew the point ##(1,0)## was a removable singularity, how could I compute:

##\lim_{(z\to 1, w\to 0)} \frac{dw}{dz}##?

I've thought about spiraling-in to the singularity numerically and maybe squeeze-out the (approx) limits but that process necessarily involves computing a numerical derivative which becomes increasingly indeterminate as I approach the singular point. I could try it I guess. Is numerically, the only way to find the limit if the solution is not known?

Perhaps I should also point out we can write the solution as:

##w(z)=\begin{cases}

1+z^{1/4}+z^{1/2}+z^{3/4} \\

1+i z^{1/4}-z^{1/2}+iz^{3/4} \\

1-iz^{1/4}-z^{1/2}+iz^{3/4} \\

1-z^{1/4}+z^{1/2}-z^{3/4}

\end{cases}

##

where we use the principal value of ##z^{1/4}## in the expressions.

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# A How to compute limit of removable singularity?

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