Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer
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:

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.