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How should I evaluate the complex limit without using L'hopital. Is that allowed to expand $\sin(z)$ in taylor when it is in the denominator? $$\large{\lim}_{z \rightarrow 3 \pi} \frac{(z-3\pi )}{(z-\pi )\sin z}$$

Robert Z
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Sagigever
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    Yes, use Taylor expansion: note that $\sin z=-\sin(z-3\pi)=-(z-3\pi)+o(z-3\pi)$ – Robert Z Jul 28 '21 at 06:44
  • @RobertZ cheers, thank you for your note, that exactly what I had need to read – Sagigever Jul 28 '21 at 06:50
  • You can not apply l'Hopital, as that would be a circular argument. The remaining fraction after applying some product and quotient limit laws has the form $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$, which is exactly the definition of the differential $f'(z_0)$ as limit of the difference quotient. See for instance https://math.stackexchange.com/questions/2118581/lhopitals-rule-and-frac-sin-xx – Lutz Lehmann Jul 28 '21 at 07:41

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