A limit two variables

How can I compute or prove that $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{\mathrm{e}^{xy}-1}{\sqrt{x^2+y^2}}=0$?

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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.

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I like that you give a proof $\underline{\textbf{directly}}$ only for $|\mathrm{e}^{xy}-1|\leq C|xy|$ for some $C>0$ in $B_1(0)$. –  bigli Aug 3 '14 at 18:35
It is wellknown that $e^{x}$ is its own derivative so that $1=\lim_{x\rightarrow0}\frac{e^{x}-1}{x}$. Then for any $C>1$ we can find a $\delta>0$ such that $\left|x\right|<\delta\Rightarrow\left|e^{x}-1\right|\leq C\left|x\right|$ –  drhab Aug 3 '14 at 18:41
Just put $z=xy$ in the inequality and you obtain the result –  Quickbeam2k1 Aug 3 '14 at 18:41
How can I use MVT in form of two variables for proof of the inequality? –  bigli Aug 3 '14 at 18:52
You don't. The inequality holds for $z\in [-1,1]$. Since $(x,y)\in B_1(0)$ you find that $xy\in(-1,1)$ and that's what suffices. –  Quickbeam2k1 Aug 3 '14 at 18:57

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:

$$\lim_{r \to 0} \frac{e^{\frac{1}{2} r^2 \sin2\theta_r} - 1}{r} = \lim_{r \to 0} \left( r \sin2\theta_r \; e^{\frac{1}{2} r^2 \sin2\theta_r} \right) = 0$$

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I was typing exactly that approach, you beat me to it.. +1 –  Ivo Terek Aug 3 '14 at 18:18
Shouldn't you be dealing with $\theta_{r}$ instead of $\theta$? The direction is not necessarily constant. –  drhab Aug 3 '14 at 18:22
@drhab like this? –  cand Aug 3 '14 at 18:42
there is no need of $\theta$ being continuous. And in fact, to show the limit in $(0,0)$ it must not be continuous –  Quickbeam2k1 Aug 3 '14 at 18:42
Who told you that $\theta(r)$ is differentiable? You better bend over to the inequality that Quickbeam2k1 is using and just drop de l'Hopital. –  drhab Aug 3 '14 at 18:43