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Thread: [SOLVED] Prove the limit of this polynomial

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

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


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

    $\displaystyle |x-2|<\delta$

    it is then true that:

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

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

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

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

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

    which simplifies to:

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

    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:

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

    which simplifies to:

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

    But how would I get that into something workable?

    Alternatively, we could go back to this:

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

    And try this:

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

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

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

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

    Now assume $\displaystyle x+2\geq0$. Then:

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

    In both cases we have this:

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

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

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

    ?

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

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

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

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

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

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

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

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

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

    So we can choose

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

    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 $\displaystyle \left| {x - 2} \right| < 1$ then $\displaystyle \left| {x + 2} \right| < 3$.
    Thus $\displaystyle \left| {f(x) - f(2)} \right| = \left| {2x^2 - 8} \right| = 2\left| {x - 2} \right|\left| {x + 2}\right| \leqslant 6\left| {x - 2} \right|$.
    Now if $\displaystyle \varepsilon > 0$ then pick $\displaystyle \delta = \min \left[ {1,\frac{\varepsilon }{6}} \right]$.
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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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