# Complex limit without L'Hospital Rule [closed]

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 $z=x+iy$.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JonMark Perry, Claude Leibovici, Leucippus, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

Note that by standard limit

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

we simply have

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

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}

• Your answer essentially use De L'Hopital, which is not allowed, doesn't it? – Taroccoesbrocco Mar 18 '18 at 8:21
• @Taroccoesbrocco No, I did not use L’Hopital. Also, I don’t know of any complex version of L’Hopital. – Jacky Chong Mar 18 '18 at 17:59

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*}