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Thread: Continuity Question

  1. #1
    Newbie Assassin0071's Avatar
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    Continuity Question

    Okay so the question is:

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

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

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

    I am letting $\displaystyle t_0 \in R$ and $\displaystyle \epsilon > 0$ then trying to find a $\displaystyle \delta > 0$ so $\displaystyle |f(t) - f(t_0)| < \epsilon$ whenever $\displaystyle |t - t_0| < \delta$ I am stuck trying to find the delta what will work. I start with $\displaystyle |f(t) - f(t_0)|$ 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 $\displaystyle f(tx)$. Note that $\displaystyle f(0x)$ is undefined, so $\displaystyle f(tx)$ can only be a continuous function of $\displaystyle t \in \mathbb{R}\setminus \{0\}$.

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

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

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

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

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