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?