proving uniformly continuous function

[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?
or maybe there is another easy way to prove this??...

 

[daon2]

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?
or maybe there is another easy way to prove this??...

My original idea was incorrect, it is easier than I thought.

\(\displaystyle \left|f(x)-f(y)\right| = \left|\dfrac{(x-y)(x+y+xy)}{(1+x)(1+y)}\right| = |(x-y)|\cdot \left|\dfrac{xy+x+y}{(1+x)(1+y)}\right|\)


Can you see how to get that, and what to do next?