I am stuck with this limit, which according to Wolfram Alpha does not exist.
$$ \begin{align} \lim_{(x,y) \to (0,0)} \frac{e^{x(y+1)} -x -1}{\|(x,y)\|} = \lim_{(x,y) \to (0,0)} \frac{e^{x(y+1)} -x -1}{\sqrt{x^2+y^2}} = \lim_{r \to 0\; \forall\theta} \frac{e^{r\cos{\theta}(r\sin{\theta}+1)} -r\cos{\theta} -1}{r} \end{align} $$
From there, I thought about using the first-degree Taylor expansion in order to get rid of the exponential (which I'm not even sure I can do).
$$ \begin{split} \lim_{r \to 0\; \forall\theta} \frac{e^{r\cos{\theta}(r\sin{\theta}+1)} -r\cos{\theta} -1}{r} &= \lim_{r \to 0\; \forall\theta} \frac{1+r\cos{\theta}(r\sin{\theta}+1) -r\cos{\theta} -1}{r} \\ &=\lim_{r \to 0\; \forall\theta} \frac{1+ r^2\cos{\theta}\sin{\theta}+r\cos{\theta} -r\cos{\theta} -1}{r}\\ & = \lim_{r \to 0\; \forall\theta} \frac{r^2\cos{\theta}\sin{\theta}}{r} \\ &= \lim_{r \to 0\; \forall\theta} {r\cos{\theta}\sin{\theta}} = 0 \end{split} $$
This is apparently wrong, but I cannot think of any other way to solve this problem, and through this method the result is clearly 0. Since I am pretty sure using Taylor's theorem here is not allowed, how else could you go about solving the limit?
By Taylor’s expansion for exponential (note that we need to expand to the second order for the presence of the $xy$ term)
$$e^{x(y+1)}=1+x(y+1)+\frac{x^2(y+1)^2}2+o(r^2)=1+x+xy+\frac{x^2}2+o(r^2)$$
then
$$\frac{e^{x(y+1)} -x -1}{\sqrt{x^2+y^2}} =\frac{1+x+xy+\frac{x^2}2+o(r^2)-x -1}{\sqrt{x^2+y^2}} =\frac{xy+\frac{x^2}2+o(r^2)}{\sqrt{x^2+y^2}} =r\cos \theta \sin \theta +r\frac{\cos^2 \theta}2+o(r)\to 0$$