Limit of $\frac{e^{xy}}{x+1}$ as it goes to $0$ I have to do the following limit: 
$$\lim_{(x,y)\rightarrow (0,0)}\frac{e^{xy}}{x+1}$$
I did that if $x = 0 \Rightarrow \lim \rightarrow 1$ but if $y = 0 \Rightarrow \lim \rightarrow$ undefined. By taking $x=y$ it gets me to the same result. Is the limit then $1$ or in what other way can I evualte it to see if it exists and if so, what's its value? 
Thanks for the help.  

Actually, if $y=0$, it's $\lim\limits_{x\to0}\dfrac1{x+1}=\color{red}1$ too.
