Using $\epsilon-\delta$ arguments, find the limit of $(e^{xy} -1)/y$ when $(x,y)\to(0,0)$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $0$.