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Thread: Problem with precise definition of the limit

  1. #1
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    Problem with precise definition of the limit

    Question :

    let $\displaystyle h(x)=x^2, x<2$
    $\displaystyle = 3, x=2$
    $\displaystyle =2, x>2$

    Show that
    lim(x-->2)h(x)does not equal to 4

    my teacher give the answer :

    for $\displaystyle 2<x<2+delta => h(x) = 2 => abs(h(x)-4)=2$

    Thus for $\displaystyle epsilon < 2, abs(h(x)-4) >= epsilon $ whenever $\displaystyle 2<x<2+delta$ no matter how small we choose
    $\displaystyle delta > 0$ => lim(x-->2)h(x)does not equal to 4


    i cannot understand the second line of the explanation. why we have to consider $\displaystyle epsilon < 2 $

    why cannot consider this : $\displaystyle abs(h(x)-4)=2 < epsilon $ thus, $\displaystyle epsilon should > 2 $ right?

    can anyone explain it to me? i really could'nt understand this!
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  2. #2
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    In order that the limit be 4, you must be able to make $\displaystyle |f(x)- 4|< \epsilon$ for any $\displaystyle \epsilon> 0$. Your teacher was pointing out that the difficulty (the fact that you can't make it "$\displaystyle <\epsilon$" happens if you take $\displaystyle \epsilon< 2$.

    The fact that it is true for some $\displaystyle \epsilon$ ($\displaystyle \epsilon> 2$) doesn't matter. It has to be true for all $\displaystyle \epsilon$.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    In order that the limit be 4, you must be able to make $\displaystyle |f(x)- 4|< \epsilon$ for any $\displaystyle \epsilon> 0$. Your teacher was pointing out that the difficulty (the fact that you can't make it "$\displaystyle <\epsilon$" happens if you take $\displaystyle \epsilon< 2$.

    The fact that it is true for some $\displaystyle \epsilon$ ($\displaystyle \epsilon> 2$) doesn't matter. It has to be true for all $\displaystyle \epsilon$.

    can you give some example, i.e. replace the epsilon and delta with a number so that i can 'see' my teacher's explanation. still confuse...

    what is the relationship between a delta and epsilon in this case? i can't see the relationship... wat the delta have to do with the epsilon value in this problem???
    anyone???
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  4. #4
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    In order to show that lim x->x_0 f(x) does not equal to L, you have to show that

    for every $\displaystyle \delta > 0 $, there exist $\displaystyle \epsilon > 0$

    thus $\displaystyle |x - x_0| < \delta$ and $\displaystyle |f(x) - L|>\epsilon$

    so in the above case

    $\displaystyle |x-2|<\delta$ ==> $\displaystyle 2-\delta<x<2+\delta$

    and thus $\displaystyle 2<x<2+\delta$ which is from the graph

    ==> h(x) = 2 ==> $\displaystyle |h(x) - L|$ = $\displaystyle |2-4|$ = $\displaystyle |-2|$ = $\displaystyle 2 >\epsilon$

    if we take $\displaystyle \epsilon = 0.0001$,thus $\displaystyle |h(x) - L|> \epsilon$ (since $\displaystyle \epsilon > 0$

    thus, it shows that lim(x-->2)h(x)does not equal to 4
    Last edited by bobey; Feb 23rd 2010 at 01:30 PM.
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