Using $\epsilon-\delta$ arguments, find the limit of $(e^{xy} -1)/y$ when $(x,y)\to(0,0)$ I'm stuck.
$$\bigg| \frac{e^{xy}-1}{y} \bigg| \leq \frac{|e^y|e^x-1|+|e^y-1|}{|y|} $$
Since $|e^\alpha-1|\leq|\alpha|e^{|\alpha|}$,
$$\bigg| \frac{e^{xy}-1}{y} \bigg| \leq \frac{|x|e^{|x|+|y|}}{|y|} + e^{|y|}$$
Any hints?
 A: Then $|e^{xy}-1|\leq|xy|e^{|xy|}$, so $\left|\dfrac{e^{xy}-1}{y}\right|\leq|x|e^{|xy|}$. We let $|x|^{2}+|y|^{2}\leq $1, then both $|x|,|y|\leq 1$ and hence $|xy|\leq 1$ and so $|x|e^{|xy|}\leq e|x|\leq e\sqrt{|x|^{2}+|y|^{2}}$. Now you let $\delta=\min\{1,\epsilon/e\}$ to finish the job.
A: If you look at the power series definition
\begin{align}
e^z = 1 + z + \dots
\end{align}
then it follows (triangle inequality) that for all $z \in \Bbb{C}$, 
\begin{align}
\left|e^z - 1\right| & \leq |z|
\end{align}
In particular, for all $(x,y) \in \Bbb{R}^2$, we have that $\left|e^{xy} - 1 \right| \leq |xy|$. So, with this, given $\epsilon > 0$, simply choose $\delta = \epsilon$. Then, for all $(x,y) \in \Bbb{R}^2$ with $0 <\lVert(x,y) \rVert < \delta$ (say the euclidean norm or maximum norm for simplicity) and $y \neq 0$, we have:
\begin{align}
\left| \dfrac{e^{xy} - 1}{y}  - 0\right| & \leq \dfrac{|xy|}{|y|} = |x| \leq \lVert (x,y)\rVert < \delta = \epsilon
\end{align}
So, this proves the limit is $0$.
A: By taking the Taylor series of $e^t$ it can be easily proved that $$|e^{xy}-1| \leq |xy|e^{|xy|}$$
Dividing by $y$,then use $\epsilon-\delta$ definition.
