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  1. #1
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    Prove discontinuity at

    Define f:\mathbb{R}\to\mathbb{R} by f(x)=8x, \ \text{if} \ x\in\mathbb{Q} and f(x)=2x^2+8 if x is irrational. Prove from the definition that f is discontinuous at 1.

    Let \epsilon >0. Then there exist \delta_1, \ \delta_2>0.

    |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}.

    Let \delta_1=\frac{\epsilon}{8}.

    |2x^2+8-10|=2|x^2-6|<\epsilon.

    Does it suffice to say that |x^2-6| can't be manipulated to some form of |x-1| now?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Prove discontinuity at

    Quote Originally Posted by dwsmith View Post
    Does it suffice to say that |x^2-6| can't be manipulated to some form of |x-1| now?
    No, it is not. Take into account that |f(x)-f(1)|=\begin{Bmatrix}8|x-1| & \mbox{ if }& x\in \mathbb{Q}\\2x^2 & \mbox{if}& x\not\in \mathbb{Q}\end{matrix}

    Now, choose for instance \epsilon =1 . If |x-1|<\epsilon/8 then 2x^2>\epsilon .
    Last edited by FernandoRevilla; July 8th 2011 at 12:19 AM. Reason: |f(x)-f(1)| instead of |f(x)-1|
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    MHF Contributor Drexel28's Avatar
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    Re: Prove discontinuity at

    Quote Originally Posted by dwsmith View Post
    Define f:\mathbb{R}\to\mathbb{R} by f(x)=8x, \ \text{if} \ x\in\mathbb{Q} and f(x)=2x^2+8 if x is irrational. Prove from the definition that f is discontinuous at 1.

    Let \epsilon >0. Then there exist \delta_1, \ \delta_2>0.

    |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}.

    Let \delta_1=\frac{\epsilon}{8}.

    |2x^2+8-10|=2|x^2-6|<\epsilon.

    Does it suffice to say that |x^2-6| can't be manipulated to some form of |x-1| now?
    I don't know how formal they want you to be but you could just say say that since there is a sequence of rationals and a sequence of irrationals converging to any point and continuous functions preserve sequences that any point $latex x_0$ where $latex f$ is continuous would have to satisfy $latex 8x_0=2x_0^2+8$--and $latex 1$ does not.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Prove discontinuity at

    Quote Originally Posted by Drexel28 View Post
    I don't know how formal they want you to be but you could just say say that since there is a sequence of rationals and a sequence of irrationals converging to any point and continuous functions preserve sequences that any point $latex x_0$ where $latex f$ is continuous would have to satisfy $latex 8x_0=2x_0^2+8$--and $latex 1$ does not.
    I suppose they don't want to use that characterization because the problem says "Prove from the definition".
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    Re: Prove discontinuity at

    Quote Originally Posted by dwsmith View Post
    Define f:\mathbb{R}\to\mathbb{R} by f(x)=8x, \ \text{if} \ x\in\mathbb{Q} and f(x)=2x^2+8 if x is irrational. Prove from the definition that f is discontinuous at 1.

    Let \epsilon >0. Then there exist \delta_1, \ \delta_2>0.

    |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}.

    Let \delta_1=\frac{\epsilon}{8}.

    |2x^2+8-10|=2|x^2-6|<\epsilon.

    Does it suffice to say that |x^2-6| can't be manipulated to some form of |x-1| now?
    Dear dwsmith,

    Using the Epsilon Delta definition of continuity, we can show that this function is discontinuous at x=1.

    Epsilon-delta definition of continuity:

    {f(x)~:~D\rightarrow\Re\mbox{ is continuous at x=c iff }\forall~\epsilon>0~\exists~\delta>0\mbox{ such that }|f(x)-f(c)|<\epsilon  \mbox{ whenever }|x-c|<\delta}.

    In this particular problem, c=1.

    f(x)=\left\{\begin{array}{cll}8x & \mbox{if}&x\in Q\\2x^{2}+8 & \mbox{if} & x\not\in Q \end{array}\right.

    |f(x)-f(1)|=\begin{Bmatrix}8|x-1| & \mbox{ if }& x\in Q\\2x^2 & \mbox{if}& x\not\in Q\end{matrix}

    Let \epsilon=0.08 and we shall try to find a \delta

    Therefore, 8|x-1|<0.08\Rightarrow |x-1|<0.01\mbox{ when }x\in Q

    2x^{2}<0.08\Rightarrow 0.8<|x-1|<1.2\mbox{ when }x\not\in Q

    We should find a \delta that satisfies for all the x values may they be rational or irrational. But this is impossible as the two intervals |x-1|<0.01\mbox{ and }0.8<|x-1|<1.2 does not have a intersection.

    That is for \epsilon=0.08 a \delta value does not exist. Therefore the function is discontinuous at x=1.

    Hope you understood my reasoning. And may there be any doubt please don't hesitate to ask.
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    Re: Prove discontinuity at

    the general strategy to prove f is NOT continuous at some point a, is to find SOME ε > 0, for which we can't find ANY δ > 0,

    for which |x - a| < δ implies |f(x) = f(a)| < ε. now let's think about this:

    8x and 2x^2 + 8 are both continuous functions. the first is "near" 8 for x "near" 1, and the second is "near" 10, for x "near 1".

    this suggests we might be able to use some 0 < ε < 2. let's try it for ε = 1.

    in order to know that 8x is within 1 of 8, we need 7 < 8x < 9, that is 7/8 < x < 9/8. that is: if |x - 1| < 1/8,

    then |8x - 8| < 1. that's pretty straight-forward. it should be clear that allowing δ > 1/8 won't work,

    then for some of the (rational) x's in (1-δ, 1+δ), f(x) will be too far away from f(1) = 8.

    now |x - 1| < 1/8 is just saying 7/8 < x < 9/8. so 49/64 < x^2 < 81/64, so 98/64 < 2x^2 < 162/64,

    so 610/64 < 2x^2 + 8 < 674/64. but 9 < 610/64, so for all x with |x - 1| < 1/8,

    for all of the rational x's, f(x) is within 1 of f(1), and this is true for NONE of the irrational x's with |x - 1| < 1/8.

    so no 0 < δ ≤ 1/8 will work (for ε = 1), and no δ > 1/8 will work, either (since SOME x's in (1-δ, 1+δ) will be

    make f(x) too far away). so, there's one ε > 0, for which it is impossible to find a suitable δ.
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