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How would one eveluate the following limit without using L'Hospital Rule $$\lim_{\Delta z\to 0}\frac{e^{\Delta z^2+2z\Delta z}-1}{\Delta z}$$ where $\Delta z=\Delta x+i\Delta y\,$ and $z=x+iy$.

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closed as off-topic by JonMark Perry, Claude Leibovici, Leucippus, José Carlos Santos, Shailesh Mar 18 '18 at 8:30

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Note that by standard limit

$$x\to 0 \qquad \frac{e^x-1}{x}\to 1$$

we simply have

$$\frac{e^{\Delta z^2+2z\Delta z}-1}{\Delta z}=\frac{e^{\Delta z^2+2z\Delta z}-1}{\Delta z^2+2z\Delta z}\frac{\Delta z^2+2z\Delta z}{\Delta z}=\frac{e^{\Delta z^2+2z\Delta z}-1}{\Delta z^2+2z\Delta z}{(\Delta z+2z)}\to2z$$

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Using Taylor expansion, we see that \begin{align} \exp((\Delta z + 2z)\Delta z) = 1+(\Delta z+2z)\Delta z+\frac{1}{2}(\Delta z+2z)^2\Delta z^2+\ldots \end{align} which means \begin{align} \frac{\exp((\Delta z + 2z)\Delta z)-1}{\Delta z} = (\Delta z+2z)+\frac{1}{2}(\Delta z+2z)^2\Delta z+\ldots \end{align} As $\Delta z\rightarrow 0$, then it follows \begin{align} \lim_{\Delta z \rightarrow 0}\frac{\exp(\Delta z^2 + 2z\Delta z)-1}{\Delta z}=2z. \end{align}

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  • $\begingroup$ Your answer essentially use De L'Hopital, which is not allowed, doesn't it? $\endgroup$ – Taroccoesbrocco Mar 18 '18 at 8:21
  • $\begingroup$ @Taroccoesbrocco No, I did not use L’Hopital. Also, I don’t know of any complex version of L’Hopital. $\endgroup$ – Jacky Chong Mar 18 '18 at 17:59
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For convenience of notation, let $h=\Delta z$.

Let $f(z)=e^{z^2}$. \begin{align*} \text{Then}\;\;&\lim_{h\to 0}\frac{e^{h^2+2zh}-1}{h}\\[4pt] =\;&\left(\frac{1}{e^{z^2}}\right)\lim_{h\to 0}\frac{\left(e^{z^2}\right)\left(e^{h^2+2zh}-1\right)}{h}\\[4pt] =\;&\left(\frac{1}{e^{z^2}}\right)\lim_{h\to 0}\frac{e^{h^2+2zh+z^2}-e^{z^2}}{h}\\[4pt] =\;&\left(\frac{1}{e^{z^2}}\right)\lim_{h\to 0}\frac{e^{(z+h)^2}-e^{z^2}}{h}\\[4pt] =\;&\left(\frac{1}{e^{z^2}}\right)\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\\[4pt] =\;&\left(\frac{1}{e^{z^2}}\right)f'(z)\\[4pt] =\;& \left(\frac{1}{e^{z^2}}\right)\left(2ze^{z^2}\right)\\[4pt] =\;&2z\\[4pt] \end{align*}

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