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Math Help - proving limit law by definition

  1. #1
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    proving limit law by definition

    there is lim_{x->x_{0}}g(x)=y_{0}, f(x)is continues in y_{0}
    point. i need to prove that lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})
    by epsilon delta definition.
    i need to follow only my books prove.help me understand it.
    the books proof:
    in order to prove lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})
    we will use the definition that for N_{\varepsilon}^{*}(f(y_{0}))
    surrounding we have N_{\delta}^{*}(x) surrounding for which every
    x\in N_{\delta}^{*}(x) f(x)\in N_{\varepsilon}^{*}(f(y_{0})).
    first step:
    f(x)is continues in y_{0} point. so lim_{x->x_{0}}f(x)=f(y_{0})
    and for N_{\varepsilon}^{*}(f(y_{0})) surrounding we have N_{\delta}^{*}(x)
    surrounding for which every
    x\in N_{\delta}^{*}(x) f(x)\in N_{\varepsilon}^{*}(f(y_{0})).
    second step:
    it is given that lim_{x->x_{0}}g(x)=y_{0} then for x\in N_{\delta_{1}}^{*}(x)
    g(x)\in N_{\varepsilon}^{*}(y_{0})
    so f((g(x))) \in N_{\varepsilon}^{*}(f(y_{0})).
    cant understand this last step why f((g(x))) \in N_{\varepsilon}^{*}(f(y_{0}))
    is the resolt of the last two steps??
    Last edited by transgalactic; August 2nd 2011 at 03:20 AM. Reason: jbjb
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  2. #2
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    Re: proving limit law by definition

    You first step concludes that if x\in N_{\delta}^{*}(x) then f(x)\in N_{\varepsilon}^{*}(f(y_{0})).

    Your second step concludes that if x\in N^*_\delta then g(x)\in N_{\varepsilon}^{*}(y_{0})

    The second step says that g(x) satisfies the condition in the first step so you replace "x" in the first step by "g(x)".
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  3. #3
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    Re: proving limit law by definition

    but we have different deltas in each limit
    the first is delta1 the other is just delta.

    and g doesnt satisfies the first one
    because its a different surroundings

    cant see wht we replace x with g(x)
    ?

    why g(x) suttisfies the first step?
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  4. #4
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    Re: proving limit law by definition

    Quote Originally Posted by transgalactic View Post
    but we have different deltas in each limit
    the first is delta1 the other is just delta.
    and g doesnt satisfies the first one
    because its a different surroundings
    cant see wht we replace x with g(x)?
    why g(x) suttisfies the first step?
    To be truthful, the book took some liberties with that proof.
    It would have been better to say, there is \delta_1>0 for the continuity and there is \delta_2>0 for the sequential limit.
    Then let \delta=\min\{\delta_1,~\delta_2\}.
    That way it insures both conditions hold.
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    Re: proving limit law by definition

    ok that is true if i choose minimal delta like you said.
    why g(x) suttisfies the first condition
    ?
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  6. #6
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    Re: proving limit law by definition

    Quote Originally Posted by transgalactic View Post
    ok that is true if i choose minimal delta like you said. why g(x) suttisfies the first condition?
    Because if \delta\le\delta_2 then N_\delta ^* (x_0 ) \subseteq N_{\delta _2 }^* (x_0 ).
    Last edited by Plato; August 2nd 2011 at 06:50 AM.
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  7. #7
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    Re: proving limit law by definition

    Quote Originally Posted by HallsofIvy View Post
    You first step concludes that if x\in N_{\delta}^{*}(x) then f(x)\in N_{\varepsilon}^{*}(f(y_{0})).

    Your second step concludes that if x\in N^*_\delta then g(x)\in N_{\varepsilon}^{*}(y_{0})

    The second step says that g(x) satisfies the condition in the first step so you replace "x" in the first step by "g(x)".
    the minimal delta thing only says that there is a smaller surrounding for which
    f(x)\in N_{\varepsilon}^{*}(f(y_{0}))
    and
    g(x)\in N_{\varepsilon}^{*}(y_{0})

    how from it we cant conlude that we can put g(x) instead of x

    ?
    Last edited by transgalactic; August 2nd 2011 at 04:47 AM. Reason: b mvb
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  8. #8
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    Re: proving limit law by definition

    Quote Originally Posted by transgalactic View Post
    the minimal delta thing only says that there is a smaller surrounding for which
    f(x)\in N_{\varepsilon}^{*}(f(y_{0}))
    and
    \color{red}g(x)\in N_{\varepsilon}^{*}(y_{0})
    how from it we cant conlude that we can put g(x) instead of x
    You don't want g(x)\in N_{\varepsilon}^{*}(y_{0})!

    You want f(g(x_0))\in N_{\varepsilon}^{*}(y_{0})
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  9. #9
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    Re: proving limit law by definition

    yes i do

    i just dont know how to get to this resolt from those two expression i am givem by taking the minimal delta

    ?
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  10. #10
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    Re: proving limit law by definition

    Quote Originally Posted by transgalactic View Post
    i just dont know how to get to this resolt from those two expression i am givem by taking the minimal delta
    Given that f is continuous at y_0 and \lim _{x \to x_0 } g(x) = y_0

    Suppose that \varepsilon  > 0.

    From continuity \left( {\exists \delta _1 >0} \right)\left[ {x \in N_{\delta _1 }^* (y_0 )\, \Rightarrow \,f(x) \in N_\varepsilon ^* \left( {f(y_0 )} \right)} \right]

    From the limit, \left( {\exists \delta _2  > 0} \right)\left[ {x \in N_{\delta _2 }^* (x_0 )\, \Rightarrow \,g(x) \in N_{\delta _1 }^* \left( {y_0 } \right)} \right].

    Let \delta  = \min \left\{ {\delta _1 ,\delta _2 } \right\}.

    Then \left[ {x \in N_\delta ^* (x_0 )\, \Rightarrow g(x) \in N_{\delta _1 }^* (y_0 )\, \Rightarrow \,\,f\left( {g(x)} \right) \in N_\varepsilon ^* \left( {f(y_0 )} \right)} \right]
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  11. #11
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    Re: proving limit law by definition

    ok i start to understand the idea
    how do i know that if we put g(x) instead of x
    i get an input of 'x' in f(x) for which this 'x' is inside the section bounded by delta1
    ?
    maybe this 'x' is out side the section bounded by delta1.
    we dont know what is g(x) we can put some x which is bounded by the minimal delta
    and get alot bigger number

    for example
    g(x)=1/x
    a=0.001
    g(a)=1000
    and get fg(a)=f(1000)
    and this 1000 could be higher the the minimal delta
    ??
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  12. #12
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    Re: proving limit law by definition

    This is my last post to this thread.
    (NOTATION: a\approx b means that a\ne b\text{ and }a\text{ is ‘close to’ }b

    From continuity if z\approx y_0 then f(z)\approx f(y_0).

    From the limit, if w\approx x_0 then g(w)\approx y_0

    Now if g(w)\approx y_0 that means f(g(w))\approx f(y_0).
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  13. #13
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    Re: proving limit law by definition

    ahhh i forgot about continuettyy
    thanks
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