Double Limit: $\lim_{a \rightarrow \infty , b \rightarrow 0^{-}} \frac{e^{ab} - 1}{b}$ I have the following double limit
$$\lim_{a \rightarrow \infty}_{b \rightarrow 0^{-}} \frac{e^{ab} - 1}{b}$$
I have a feeling this is undefined, since
$$\lim_{a \rightarrow \infty} \lim_{b \rightarrow 0^{-}} \frac{e^{ab} - 1}{b}$$ is undefined, but
$$\lim_{b \rightarrow 0^{-}} \lim_{a \rightarrow \infty} \frac{e^{ab} - 1}{b} \rightarrow \infty$$
Are any of these right? 
 A: We may assume $a>0$ since we are going to let $a$ tend to $\infty$.For each fixed $a>0$ we have
$\lim_{b\to0^{-}}\dfrac{e^{ab}-1}{b}=\lim_{b\to0^{-}}\dfrac{e^{ab}-(e^{a})^{0}}{b}=\frac{d}{db}(e^{ab})|_{b=0^{-}}=a$
This is the case for each fixed $a$. So letting $a$ tend to $\infty$ makes the above tend to $\infty$. So the iterated limit in this order is infinite.
If we fix $b<0$ then
$\lim_{a\to\infty}\dfrac{e^{ab}-1}{b}=\dfrac{1}{-b}$.
Since $-b>0$ for $b<0$ then letting $b$ tend to $0$ from the left makes the above tend to infinity. This is true no matter where we fix $b$ to take the limit in $a$ as long as $b<0$.
We wish to check the double limit. We first make note of a few things. First fix $a>0$. There is $\delta>0$ such that if $b<0$ and $-b<\delta$ then $\dfrac{e^{ab}-1}{b}>a-1$ (by the first limit we showed). Now suppose $a_{1}>a$. Then for any $b$ such that $b<0$ and $-b<\delta$ we have:
$a_{1}b<ab$
$e^{a_{1}b}<e^{ab}$ 
$e^{a_{1}b}-1<e^{ab}-1$ 
$\dfrac{e^{a_{1}b}-1}{b}>\dfrac{e^{ab}-1}{b}>a-1$
So once we choose $\delta$ for some $a>0$ the lower bound we get holds for all larger $a$.
Now we proceed with the proof. Fix some $a_{0}$ such that $a_{0}>N+1$. Then choose $\delta>0$ such that if $b<0$ and $-b<\delta$ then $\dfrac{e^{a_{0}b}-1}{b}>a_{0}-1$. It holds for all larger $a$ as shown above. But by our choice of $a_{0}$ we have the following:
$\dfrac{e^{ab}-1}{b}>a_{0}-1>N$ for all $a\ge a_{0}$ (if we remain in the delta region set chosen before). Hence if we remain in the set $\{(a,b)|a\ge a_{0}\,\,and\,\,-b<\delta\,\,and\,\,b<0\}$ then $\dfrac{e^{ab}-1}{b}>N$. This shows for each $N\in\mathbb{N}$ we can find $N_{0}$ and $\delta$ such that if $a>N_{0}$ and $|b|=-b<\delta$ then $\frac{e^{ab}-1}{b}>N$
Since $N$ was arbitrary we can say that the double limit is infinity.
