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Math Help - Continuity Question

  1. #1
    Newbie Assassin0071's Avatar
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    Continuity Question

    Okay so the question is:

    Let f:R^2 \rightarrow R by
    f(x) = \frac{x_1^2x_2}{x_1^4+x_2^2} for x \not= 0

    Prove that for each x \in R, f(tx) is a continuous function of t \in R

    ([MTEXR[/MTEX is the real numbers, I'm not sure how to get it to look right).

    I am letting t_0 \in R and \epsilon > 0 then trying to find a \delta > 0 so |f(t) - f(t_0)| < \epsilon whenever |t - t_0| < \delta I am stuck trying to find the delta what will work. I start with |f(t) - f(t_0)| and try and simplify but I am not sure how to get to the point where I can determine delta. Any help appreciated.
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  2. #2
    MHF Contributor
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    Re: Continuity Question

    I think what you want is: \mathbb{R}

    Anyway, let's calculate f(tx). Note that f(0x) is undefined, so f(tx) can only be a continuous function of t \in \mathbb{R}\setminus \{0\}.

    f(tx) = \dfrac{(tx_1)^2(tx_2)}{(tx_1)^4 + (tx_2)^2} = \dfrac{tx_1^2x_2}{t^2x_1^4 + x_2^2} = \dfrac{tx_1^{-2}x_2}{t^2+(x_1^{-2}x_2)^2}

    Note: The final simplification is only possible if x_1\neq 0. If x_1=0, then f(tx)=0 is a constant function and obviously continuous. Similarly, if x_2=0, you also have a constant function, which is obviously continuous. So, assume x_1\neq 0 and x_2\neq 0.

    Since f(-tx) = -f(tx), the function is odd. So, you only need to consider continuity for t>0 and by symmetry, you will find continuity for all t\in \mathbb{R}\setminus \{0\}.

    Now, given x\in \mathbb{R}^2 \setminus \{(0,0)\}, t_0\in \mathbb{R}\setminus \{0\}, and \varepsilon>0, you want to find \delta>0 that works. First of all, we are going to want to limit the possible range of \delta. We know that the function is not continuous at t=0, so 0 < \delta < |t_0|.

    Does this help at all?
    Last edited by SlipEternal; October 20th 2013 at 07:39 PM.
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