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Prove a function is continuos


Feb 3, 2013
Prove that
e^{-1/|x-y|},x\neq y\\
is continuous on $\mathbb{R}^2$.

Can I conclude since $e^{-1/t}$ is continuous then for $x$,$y\in \mathbb{R}^2$, $x\neq y$,$e^{-1/|x-y|}$ is continuous on $\mathbb{R}^2$ since here $t=|x-y|$? And how to prove it when $x=y$, using the $\delta- \varepsilon$ way? Thank you a lot!


Feb 1, 2012
Re: Prove a function is continuous

We have to show that the function is continuous at each point of $\Bbb R^2$. Your argument threats the case of points of the form $(x,y),x\neq y$. So now, fix $x_0$. We have to show that $\lim_{(x,y)\to (x_0,x_0)}=0$.

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each non-negative real number $u$.
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
Re: Prove a function is continuous

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each real number $u$.
Surely is a typo, I suppose you meant for each real number $u$ close to $0$.

Another way: denoting $\Delta=\{(t,t):t\in\mathbb{R}\}$ (diagonal of $\mathbb{R}^2$), in a neighborhood $V$ of $(0,0)$ we have:

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}e^{-\dfrac{1}{|x-y|}}=e^{-\infty}=0$

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}0=0$

In a finite partition of $V$ all the limits coincide, so $\displaystyle\lim_{(x,y)\to (x_0,x_0)}f(x,y)=0$ .


Feb 3, 2013
The definition of continuity for a function is {lim x-> a} f(x) = f(a). For this function we only need to worry about when |x - y| -> 0, since the exponential function is continuous everywhere. So in other words we need to show that
{lim |x - y|-> 0} f(x,y) = f(x,x) = f(y,y)
Now f(x,x) = f(y,y) = 0 so we just need to show that
{lim |x - y| -> 0} f(x,y) = 0.
Since |x - y| > 0, -1/|x - y| -> infinity so f(x,y) -> 0 and we're done.

Is a proof like this correct?