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Math Help - [SOLVED] Prove the limit of this polynomial

  1. #1
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    [SOLVED] Prove the limit of this polynomial

    Prove that for f:\mathbb{R}\to\mathbb{R} defined as f(x)=2x^2+8, f has a limit at 2.


    Obviously the limit is 16. So we need to show that for each \epsilon>0, there exists \delta>0 such that for each x satisfying:

    |x-2|<\delta

    it is then true that:

    |(2x^2+8)-16|<\epsilon

    So let's get busy. The last bit simplifies to:

    |x^2-4|<\frac{\epsilon}{2}

    I can only guess what to do next. I could try this:

    -\frac{\epsilon}{2}<x^2-4<\frac{\epsilon}{2}

    which simplifies to:

    4-\frac{\epsilon}{2}<x^2<4+\frac{\epsilon}{2}

    But this is a little strange. I want to take the square root, but the left side could be negative, so I can't. So, maybe I could do:

    max\{0,4-\frac{\epsilon}{2}\}<x^2<4+\frac{\epsilon}{2}

    which simplifies to:

    \sqrt{max\{0,4-\frac{\epsilon}{2}\}}<|x|<\sqrt{4+\frac{\epsilon}{  2}}

    But how would I get that into something workable?

    Alternatively, we could go back to this:

    -\frac{\epsilon}{2}<x^2-4<\frac{\epsilon}{2}

    And try this:

    -\frac{\epsilon}{2}<(x-2)(x+2)<\frac{\epsilon}{2}

    But then we have a problem of negatives. So let's assume x+2<0. Then:

    -\frac{\epsilon}{2(x+2)}>(x-2)>\frac{\epsilon}{2(x+2)}

    |x-2|<-\frac{\epsilon}{2(x+2)}

    Now assume x+2\geq0. Then:

    |x-2|<\frac{\epsilon}{2(x+2)}

    In both cases we have this:

    |x-2|<\frac{\epsilon}{2|x+2|}

    But still we have a problem. How do we find \delta from this:

    \delta\leq\frac{\epsilon}{2|x+2|}

    ?

    Or am I going about this entirely the wrong way?

    Any help would be much appreciated!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    |x^2-4|<\frac{\epsilon}{2}

    I can only guess what to do next. I could try this:
    ...
    Let's do

    \lvert x^2-4\rvert<\frac\epsilon2

    \Rightarrow\lvert(x-2)(x+2)\rvert<\frac\epsilon2

    \Rightarrow\lvert x-2\rvert\lvert x+2\rvert<\frac\epsilon2

    Now, for 0<\lvert x-2\rvert<1 (i.e., if we only consider \delta\leq1), we have 3<x+2<5 and

    \lvert x-2\rvert\lvert x+2\rvert<\frac\epsilon2

    \Leftarrow5\lvert x-2\rvert<\frac\epsilon2

    \Leftrightarrow\lvert x-2\rvert<\frac\epsilon{10}

    So we can choose

    \delta=\min\left\{1,\frac\epsilon{10}\right\}.

    Now, working through backwards, you can write a clean proof.
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  3. #3
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    Here is a way I like to start these. Suppose that \left| {x - 2} \right| < 1 then \left| {x + 2} \right| < 3.
    Thus \left| {f(x) - f(2)} \right| = \left| {2x^2  - 8} \right| = 2\left| {x - 2} \right|\left| {x + 2}\right| \leqslant 6\left| {x - 2} \right|.
    Now if \varepsilon  > 0 then pick \delta  = \min \left[ {1,\frac{\varepsilon }{6}} \right].
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  4. #4
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    Thanks a million, guys! It's so simple... I must be getting old.
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