Multi variable limit $\lim\limits_ {(x, y) \to (0, 0)} \frac {e^{xy} − 1} y$ 
I am looking at :
  $$\lim\limits_ {(x, y) \to (0, 0)} \frac {e^{xy} − 1} y$$

and choosing the path $y = 0$, the limit becomes:
$$\lim\limits_ {(x, 0) \to (0, 0)} \frac {1 − 1} 0$$
On an answer key, this limit evaluates to 0 but I don't know why because 0/0 should be undefined.
 A: The path ${y=0}$ is not in the domain of your function. 
For $y\ne 0$, we have $$e^{xy}=1+xy+\frac {x^2y^2}{2!}+.....$$
$$ e^{xy}-1 = xy+\frac {x^2y^2}{2!}+.....$$
$$ \frac{e^{xy}-1}{y} = x+\frac {x^2y}{2!}+.....$$
$$ lim_{(x,y)\to (0,0)} \frac{e^{xy}-1}{y} =$$
$$ lim_{(x,y)\to (0,0)} x+\frac {x^2y}{2!}+.....   = {0} $$
A: 
I thought it might be instructive to present an approach that relies only on a pair of elementary inequalities and the squeeze theorem.  To that end, we proceed.


In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that
$$1+x\le e^{x}\le \frac{1}{1-x}\tag 1$$
for $x<1$.  Using $(1)$, we have for $xy<1$
$$x\le \frac{e^{xy}-1}{y}\le \frac{x}{1-xy}$$
whence the squeeze theorem guarantees that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{e^{xy}-1}{y}=0}$$
And we are done!
A: To evaluate a limit $f(z)/g(z)$ as $z\to 0$ you generally cannot just take $f(0)/g(0).$ For example if $f(z)=g(z)=z$ for every $z,$ then $f(z)/g(z)=1$ when $z\ne 0$ but $f(0)/g(0)$ does not exist. 
It is  tacitly assumed that if $g(0)=0$ then the limit of $f(z)/g(z)$ as $z\to 0 $ is evaluated as the limit as $z\to 0$ through non-zero values. 
(i). For $y\ne 0\ne x$ we have $$\frac {e^{xy}-1}{y}=x\cdot \frac {e^{xy}-1}{xy}=x\cdot \frac {f(z)-f(0)}{z-0}$$ where $z=xy$ and $f(z)=e^z.$ As $(x,y)\to (0,0)$ with $y\ne 0\ne x$ we have   $z\to 0$ so $$\frac {f(z)-f(0)}{z-0}\to f'(0)=e^0=1.$$ And we also have $x\to 0,$ so $x\cdot \frac {f(z)-f(0)}{z-0}\to 0\cdot 1=0.$
(ii). For $y\ne 0$ and $x=0$ we have $\frac {e^{xy}-1}{y}=\frac {e^0-1}{y}=0.$
By (i) and (ii), if $(x,y)\to (0,0)$ with $y\ne 0$ then  $\frac{e^{xy}-1}{y}\to 0.$
A: $$\lim\limits_ {(x, y) \to (0, 0)} \frac {e^{xy} − 1} y= \lim\limits_ {(x, y) \to (0, 0)} x\cdot\lim\limits_ {(x, y) \to (0, 0)}\frac {e^{xy} − 1}{ xy}= \lim\limits_ {{(x, y) \to (0, 0)}} x\cdot\lim\limits_ {h \to 0}\frac {e^{h} − 1}{ h}=0*1=0$$
A: Note that
$$\frac {e^{xy} − 1} y=x \cdot \frac {e^{xy} − 1} {xy}\to 0\cdot1=0$$
indeed
$$e^{xy}=1+xy+o(\rho^2)\implies \frac {e^{xy} − 1} {xy}=\frac {1+xy+o(\rho^2) − 1} {xy}=\frac {xy+o(\rho^2)} {xy}=1+o(1)\to1$$
