TITLE: A limit two variables
QUESTION [2 upvotes]: How can I compute or prove that $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{\mathrm{e}^{xy}-1}{\sqrt{x^2+y^2}}=0$?

REPLY [5 votes]: For small $z\in [-1,1]$ there holds $|e^z-1| \leq C|z|$ for some constant $C>0$. Hence for $(x,y)\in B_1(0)$ we may estimate
$$ \left|\frac{e^{xy}-1}{\sqrt{x^2+y^2}}\right| \leq C \frac{|xy|}{\sqrt{x^2+y^2}}$$
Now observe that $2ab\leq a^2+b^2$ and you find
$$ C \frac{|xy|}{\sqrt{x^2+y^2}} \leq \frac{C}{2} \sqrt{x^2+y^2}\rightarrow 0$$ as $(x,y)\to (0,0)$
To show the inequality $|e^z-1|\leq C|z|$ note that $1=e^0$ and use the mean value theorem.
For $z=0$ the inequality is clear for any $C>0$. For $z\in (0,1]$ we find with the MVT and some $\xi\in(0,z)$ that
$$ e^z-1 = e^z \xi \leq e^1 z=e z$$
The estimate on $[-1,0)$ is proven analogously (or with the functional relation of the exponential) and we see that the constant can be chosen such that $C=e$ which is even optimal here.

REPLY [1 votes]: Suppose that $\theta_r$ is continuous and differentiable everywhere on $\mathbb R$. This allows the substitution $$x = r\cos\theta_r \\ y = r\sin\theta_r$$
Then $(x,y) \mapsto (0,0) = r \mapsto 0$, so:
$$\begin{align}
\lim_{(x,y) \to (0,0)} \frac{e^{xy}-1}{\sqrt{x^2+y^2}}
& = \lim_{r \to 0} \frac{e^{r^2 \sin\theta_r \cos\theta_r} - 1}{\sqrt{r^2 \cos^2\theta_r + r^2 \sin^2\theta_r}} \\
& = \lim_{r \to 0} \frac{e^{\frac{1}{2} r^2 \sin2\theta_r} - 1}{r}
\end{align}$$
This is a limit of type $\frac{0}{0}$, so we can use L'Hopital:
$$\lim_{r \to 0} \frac{e^{\frac{1}{2} r^2 \sin2\theta_r} - 1}{r}
= \lim_{r \to 0} \left( r \sin2\theta_r \; e^{\frac{1}{2} r^2 \sin2\theta_r} \right)
= 0$$
{"set_name": "stack_exchange", "score": 2, "question_id": 886495}