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Thread: delta epsilon

  1. #1
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    delta epsilon

    USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
    lim -3x+7=10
    x-(-1)
    Last edited by tumi; Feb 8th 2009 at 04:42 AM.
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  2. #2
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    Quote Originally Posted by tumi View Post
    USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
    lim -3x+7=10
    x-0
    Find any post by this guy: http://www.mathhelpforum.com/math-he.../krizalid.html

    eg. http://www.mathhelpforum.com/math-he...evolution.html

    Click on the obvious link in his signature.
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  3. #3
    Super Member Showcase_22's Avatar
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    Doesn't -3x+7 \rightarrow+7?

    Since -3x+7 is a continuous function it's limit is equal to when x=0. This gives +7.
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  4. #4
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    As Showcase_22 says, you cannot prove \lim_{x \rightarrow 0} -3x+ 7= 10- it isn't true!
    What you can prove is either \lim_{x\rightarrow 0} -3x+ 7= 7 or \lim_{x\rightarrow -1} -3x+ 7= 10. Is your problem either of those?
    Last edited by HallsofIvy; Feb 7th 2009 at 11:00 AM.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post

    Click on the obvious link in his signature.
    Well in this section that thread is stickied.
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  6. #6
    Eater of Worlds
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    Yes indeed. There is a typo or something in your statement.

    Let's make up something similar and go with that as an example.

    \lim_{x\to 2}[2x-7]=-3

    |(2x-7)+3|=|2x-4|=2|x-2|<{\epsilon} \;\ if \;\ |x-2|<\frac{\epsilon}{2}, \;\ so \;\ {\delta}=\frac{\epsilon}{2}

    Here's another. I am just making an example up to illustrate to you as a tutorial. Hope it helps.

    \lim_{x\to 4}\sqrt{x}=2

    |\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\fr  ac{|x-4|}{\sqrt{x}+2}

    Now, if we pull the reins back on delta and restrict it so that {\delta}\leq 4, then:

    |x-4|<4
    0<x<8
    0<x<\sqrt{8}
    2<\sqrt{x}+2<\sqrt{8}+2
    \frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8}  +2}
    \frac{1}{\sqrt{x}+2}<\frac{1}{2}
    \frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|

    Therefore, thus, and all that:

    |\sqrt{x}-2|<{\epsilon} \;\ if \;\ \frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ if \;\ |x-4|<2{\epsilon}, \;\ so \;\ {\delta}=min(2{\epsilon},4)

    There.....does that help finish the one you have assuming you fix the typo or what not?.
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  7. #7
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    my apologies

    Sorry i made a mistake. [tex]\lim_{x\rightarrow -1} -3x+ 7= 10 is what i need to prove.can you please help me.
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  8. #8
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    At least read the above posts.
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