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Thread: proving uniformly continuous function

  1. #1
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    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?
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  2. #2
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    Re: proving uniformly continuous function

    Quote Originally Posted by orir View Post
    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
    Spoiler:

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


    Estimating some things.
    Spoiler:

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


    Hence
    Spoiler:

    $\displaystyle \left|\frac{x^2}{1+x}-\frac{y^2}{1+y}\right| \leq |x-y| $ so you may as well choose $\displaystyle \delta = \epsilon$.
    Last edited by Gusbob; Apr 18th 2013 at 10:00 PM.
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