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Math Help - Find delta given epsilon

  1. #1
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    Find delta given epsilon

    Could you help me with the problem?

    Find delta using the definition of limits, given epsilon = 0,25

    lim 1 / (2-x) = -1/3
    x->5

    Answer should be delta = 1
    How can I get it?

    Thanks.
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  2. #2
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    Do you really need to apply a \delta - \epsilon proof here?
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  3. #3
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    Quote Originally Posted by sodk View Post
    Could you help me with the problem?

    Find delta using the definition of limits, given epsilon = 0,25

    lim 1 / (2-x) = -1/3
    x->5

    Answer should be delta = 1
    How can I get it?

    Thanks.
    No, the answer is not 1- but 1 is a perfectly valid answer.

    The definition of "limit" says that, given any \epsilon> 0, there exist \delta> 0 such that if |x- 5|< \delta then \left|\frac{1}{2- x}- (-\frac{1}{3}\right|< \epsilon.

    Here, you are given that \epsilon= 0,25 so we want to have \left|\frac{1}{2-x}+ \frac{1}{3}\right|< 0.25

    Go ahead and add the fractions on the left:
    \frac{3}{3(2-x)}+ \frac{2- x}{3(2-x)}=\frac{5- x}{3(2-x)}

    So we want \left|\frac{5-x}{3(2-x)}\right|< .25.

    That is the same as \left|5-x|\right|< \frac{3}{4}|x-2|

    Let's say that |5- x|< 1. Then -1< x- 5< 1 so 4< x< 6. then 2< x- 2< 4 so 2< |x-2|< 4. Certainly, \frac{3}{4}(2)= \frac{3}{2}< \frac{3}{4}|x-2| then, so if we can make |x- 5|< \frac{3}{2} we are done. Since we are already assuming |x- 5|< 1, and 1< \frac{3}{2}, it is sufficient to make |x- 5|< 1. That is, we can take \delta= 1.

    Notice that this is not saying that \delta has to be equal to 1. Obviously any smaller number would work and there might be some larger value that would work. But \delta= 1 is sufficient.

    Had I said "Let's say that |5- x|<" some other number, I would have gotten an answer other than 1, but equally valid.
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