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$$\lim _{z\to 0}\frac{d^2}{dz^2}\left(\frac{e^zz}{\sin z}\right)=\lim _{z\to 0}\left(\frac{d}{dz}\left(\frac{e^z\:z+e^z}{\sin z}-\frac{ze^z\cos z}{\sin^2 z}\right)\right)$$

Hi, can someone help me calculate the limit? The solution the test had is using taylor series, but to be honest, I would have never thought of it and always did using the derivative. for some reason, I cant reach a correct answer without using taylor. Much help will be appreciated.

user
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LearningToCode
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    @ThomasAndrews ... that's why he said "without Taylor". – GEdgar Jul 14 '23 at 13:57
  • is it possibly that it is impossible without taylor? because it is really weird, never had a limit like that. if it really is impossible, I guess taylor is the way, althought I do not like it. – LearningToCode Jul 14 '23 at 14:01
  • You just need to find $\lim\limits_{z\to0}\csc^3(z)(1+2z-\cos(2z)-(z+1)\sin(2z))$ – Тyma Gaidash Jul 14 '23 at 14:05
  • problem is, it probably requires taylor, since limit of $\frac 1 {sin(z)}$ is not defined and such, I dont know how to explain, but without taylor it is impossible then, right? – LearningToCode Jul 14 '23 at 14:08

2 Answers2

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As Thomas said, the easy way to do this is with Taylor.

Without Taylor?

$$ \frac{d^2}{dz^2} f(z) = {\frac {{2{\rm e}^{z}} \left( \left( \sin \left( z \right) \right) ^{2}z-\sin \left( z \right) \cos \left( z \right) z+ \left( \cos \left( z \right) \right) ^{2}z+ \left( \sin \left( z \right) \right) ^{2}-\sin \left( z \right) \cos \left( z \right) \right) }{ \left( \sin \left( z \right) \right) ^{3}}} $$ The limit of this as $z \to 0$ is indeterminate of form $0/0$. Attempt l'Hopital's rule $$ -{\frac {2\,{{\rm e}^{z}} \left( \cos \left( z \right) z-2\,\sin \left( z \right) z-3\,\sin \left( z \right) \right) }{3\,\sin \left( z \right) \cos \left( z \right) }} $$ This limit at $z=0$ is still indeterminate of form $0/0$. Do it again $$ {\frac {2\,{{\rm e}^{z}} \left( 2\,\sin \left( z \right) z-\cos \left( z \right) z+3\,\sin \left( z \right) \right) +2\,{{\rm e}^{z} } \left( 2\,\cos \left( z \right) z+2\,\sin \left( z \right) +\sin \left( z \right) z+2\,\cos \left( z \right) \right) }{3\, \left( \cos \left( z \right) \right) ^{2}-3\, \left( \sin \left( z \right) \right) ^{2}}} $$ Now we can plug in $z=0$ to get the answer $$ \frac{4}{3} $$

GEdgar
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  • oh, I dont think we learnt lhopital at complex, but I guess if I have something like that at test, I will really use it, i dont think there will be a problem. Thank you very much!! It helped me alot :) – LearningToCode Jul 14 '23 at 14:12
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As an alternative, without l'Hospital, just using standard limits we have

$$\frac{d^2}{dz^2} f(z) = \frac {{2{\rm e}^{z}} \left( z\sin^2 z-z\sin z \cos z + z\cos^2 z+ \sin^2 z-\sin z \cos z \right) }{\sin^3 z}=$$

$$=2e^z \left(\frac {z}{\sin z}+(\cos z - \sin z)\frac {z\cos z - \sin z}{\sin^3 z}\right)$$

with $e^z \to 1$ and $\frac {z}{\sin z}\to 1$, $(\cos z - \sin z)\to 1$ and

$$ \lim _{z\to 0}\frac {z\cos z - \sin z}{\sin^3 z}=-\frac 1 3 $$

indeed using these results we have

$$\small \frac {z\cos z - \sin z}{\sin^3 z}=\frac {z\cos z-z+z - \sin z}{\sin^3 z}=\frac{z^3}{\sin^3 z}\left(\frac{\cos z-1}{z^2}+\frac{z-\sin z}{z^3}\right)\to 1\cdot \left(-\frac12+\frac16\right)=-\frac13$$

then finally

$$2e^z \left(\frac {z}{\sin z}+(\cos z - \sin z)\frac {z\cos z - \sin z}{\sin^3 z}\right)\to 2\cdot \left(1-\frac13\right)=\frac43$$

user
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  • Oh dam, Very good one, it uses lots of thinking skills ( multi and divide z and such ), great ideas, I will learn this carefully. Huge thanks :) – LearningToCode Jul 14 '23 at 16:06
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    @LearningToCode Yes! If you are not familiar with Taylor, standard limits can be very useful to obtain simpler expression and often the result directly. You are welcome. Bye – user Jul 14 '23 at 16:17