proving uniformly continuous function

• Apr 18th 2013, 01:56 PM
orir
proving uniformly continuous function
i need to show that $f(x)=\frac{x^{2}}{1+x}$ is uniformly continuous at $[0,\infty)$.
i was trying to show that $\forall(\varepsilon>0)\exists(\delta>0)\forall|x-y|<\delta$ maintains $|\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq\varepsilon
$
. $\:$what i was getting is that in the given range - $|\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq|x-y|\cdot[x+y+xy]\leq\delta\cdot[x+y+xy]$, and then i was stuck... how can i proceed from here?
• Apr 18th 2013, 10:56 PM
Gusbob
Re: proving uniformly continuous function
Quote:

Originally Posted by orir
i need to show that $f(x)=\frac{x^{2}}{1+x}$ is uniformly continuous at $[0,\infty)$.
i was trying to show that $\forall(\varepsilon>0)\exists(\delta>0)\forall|x-y|<\delta$ maintains $|\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq\varepsilon
$
. $\:$what i was getting is that in the given range - $|\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq|x-y|\cdot[x+y+xy]\leq\delta\cdot[x+y+xy]$, and then i was stuck... how can i proceed from here?

I believe you eliminated the denominator too early, your current approach won't work. Below is a solution, I have spoilered it in parts so you can jump on in whenever you wish.

Keeping the denominator
Spoiler:

$\left|\frac{x^2}{1+x}-\frac{y^2}{1+y}\right| = \left| \frac{(x-y)(x+y+xy)}{1+x+y+xy}\right| \leq |x-y|\cdot \left| \frac{x+y+xy}{1+x+y+xy}\right|$

Estimating some things.
Spoiler:

Since $x+y+xy < 1+x+y+xy$, you get $\left| \frac{x+y+xy}{1+x+y+xy}\right|\leq 1$.

Hence
Spoiler:

$\left|\frac{x^2}{1+x}-\frac{y^2}{1+y}\right| \leq |x-y|$ so you may as well choose $\delta = \epsilon$.