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Math Help - Need help finding delta....

  1. #1
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    Need help finding delta....

    Problem:

    let f(x)=sqrt(19-x) and lim x-->10 f(x)=3 . Use the graph (or table) to find a value delta whenever epsilon=.5

    Can someone please help me with this problem? Thanks
    I don't even know where to begin...
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    You are looking for a value of \delta such that whenever |x-10|<\delta then |f(x)-f(10)|<\epsilon.

    Where do you get stuck?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by johnny4lsu View Post
    Problem:

    let f(x)=sqrt(19-x) and lim x-->10 f(x)=3 . Use the graph (or table) to find a value delta whenever epsilon=.5

    Can someone please help me with this problem? Thanks
    I don't even know where to begin...

    This is how I do this

    But, your problem says to find the value by lookind at the graph, which is much easier. I'll talk about that in a minute.
    f(x)=\sqrt{19-x}

    \lim_{x\to{10}}\sqrt{19-x}=3

    We wish to find an \delta such that |\sqrt{19-x}-3|<0.5 whenever |x-10|<\delta

    So here we just solve the double inequality

    -0.5<\sqrt{19-x}-3<0.5

    2.5<\sqrt{19-x}<3.5
    squaring (note that the direction of the signs is preserved)

    6.25<19-x<12.25

    -12.75<-x<-6.75

    12.75>x>6.75

    So now we need to look at
    |x-10|<\delta

    So we choose \delta=2.75


    Analyling limits graphically, although imprecise, is easy.

    1.Graph the function

    2. graph the lines y=3, y=3.5, and y=2.5

    3. where the lines y=3.5 and y=2.5 intersect the graph of the function, draw vertical lines.

    x=10 will be in the interval where these lines cross. Approximate delta by choosing delta just a little bit smaller than the absolute value of the difference between 10 and the aproximate value of x where the vertical line closest to x=10 crosses the x-axis. this is your delta.
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  4. #4
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    Smile

    thanks a bunch. that way of doing it makes much more sense to me. Thanks again!
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