proving uniformly continuous function

i need to show that $\displaystyle f(x)=\frac{x^{2}}{1+x}$ is uniformly continuous at $\displaystyle [0,\infty) $.

i was trying to show that $\displaystyle \forall(\varepsilon>0)\exists(\delta>0)\forall|x-y|<\delta$ maintains $\displaystyle |\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq\varepsilon

$. $\displaystyle \:$what i was getting is that in the given range - $\displaystyle |\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq|x-y|\cdot[x+y+xy]\leq\delta\cdot[x+y+xy]$, and then i was stuck... how can i proceed from here?

Re: proving uniformly continuous function

Quote:

Originally Posted by

**orir** i need to show that $\displaystyle f(x)=\frac{x^{2}}{1+x}$ is uniformly continuous at $\displaystyle [0,\infty) $.

i was trying to show that $\displaystyle \forall(\varepsilon>0)\exists(\delta>0)\forall|x-y|<\delta$ maintains $\displaystyle |\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq\varepsilon

$. $\displaystyle \:$what i was getting is that in the given range - $\displaystyle |\frac{x^{2}}{1+x}-\frac{y^{2}}{1+y}|\leq|x-y|\cdot[x+y+xy]\leq\delta\cdot[x+y+xy]$, 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

Estimating some things.

Hence