# Limit of two-variable function: ${\lim_{(x,y) \to (0,0)} \frac{e^{x(y+1)} -x -1}{\|(x,y)\|}}$

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?

• Let $f$ be $\displaystyle (x,y)\mathop{\mapsto}^{\mathbb R^2}e^{x(y+1)}$. One the one hand $f$ is differentiable (by the standard theorems). On the other hand, and by definition, it is differentiable if, and only if, $$\lim \limits_{(x,y)\to (0,0)}\Bigg(\dfrac{f(x,y) - [\overbrace{f(0,0)}^{=1} + \overbrace{f_{x}(0,0)(x-0)}^{=x} + \overbrace{f_{y}(0,0)(y-0)}^{=0}]}{\sqrt{x^2 + y^2}}\Bigg)=0.$$ So if nothing better you can just go ahead and use the proof of the standard theorems. Mar 18, 2018 at 20:17
• Please clarify: are you not allowed to use Taylor’s expansion or you think that it is not applicable?
– user
Mar 18, 2018 at 20:43
• @gimusi The question is clearly stated.
– Did
Mar 18, 2018 at 20:46
– user
Mar 18, 2018 at 20:47
• Sorry, it's that I thought it was not applicable. However, after thinking about it some more, I don't see why it wouldn't be. Mar 18, 2018 at 23:10

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

$$\frac{e^{x(y+1)} -x -1}{\sqrt{x^2+y^2}} =\frac{1+x+xy+\frac{x^2}2+o(r^2)-x -1}{\sqrt{x^2+y^2}} =\frac{xy+\frac{x^2}2+o(r^2)}{\sqrt{x^2+y^2}} =r\cos \theta \sin \theta +r\frac{\cos^2 \theta}2+o(r)\to 0$$

• "Note that ... $\frac{ x}{\sqrt{x^2+y^2}} \to 0$" Well, no. This ratio has no limit when $(x,y)\to(0,0)$.
– Did
Mar 18, 2018 at 19:31
• @Did that’s a good and useful observation, thanks I fix
– user
Mar 18, 2018 at 19:35