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Math Help - How do I prove this limit using the δ , ε method of proof?

  1. #1
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    How do I prove this limit using the δ , ε method of proof?

    Hi, can somebody please help me how to prove this limit:

    lim x->2 (2x + 3) / (x - 1)

    using the δ , ε method of proof?

    I know that:

    if 0 < | x - 2 | < δ then | ( (2x + 3) / (x - 1) ) - 7 | < ε

    but i'm not quite sure what to do after that

    Thanks!
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  2. #2
    o_O
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    \lim_{x \to 2} \frac{2x + 3}{x-1} = 7

    By definition:
    \forall \epsilon > 0, \ \exists \delta > 0 such that whenever 0 < |x - 2| < \delta then it follows that \left| \frac{2x+3}{x - 1} - 7 \right| < \epsilon

    In these questions, you always fudge the last expression to get |x - 2| < \text{something} in terms of \epsilon in which you set \delta to. So:

    \begin{array}{rcl} \left| \displaystyle \frac{2x + 3}{x-1} - \frac{7(x-1)}{x-1} \right| & < & \epsilon \qquad \text{Found common denominator}\\ & & \\ \left| \displaystyle \frac{2x + 3 - 7x + 7}{x-1} \right| & < & \epsilon \\ & & \\ \left| \displaystyle \frac{-5x + 10}{x - 1} \right| & < & \epsilon \\ & & \\ |-5| \displaystyle \frac{|x-2|}{{\color{red}|x-1|}} & < & \epsilon \end{array}

    So the problem is the |x - 1|. To deal with this, we can assume that we're dealing with relatively small \delta values so we can say \delta \leq \frac{1}{2} (Normally we use \delta \leq 1 but we run into a bit of a problem so we choose a different value). So:
    |x - 2| < \delta \leq \frac{1}{2}  \ \ \Rightarrow \ \ -\frac{1}{2} \ < \ x - 2 \ < \ \frac{1}{2}

    Adding +1 to all 3 sides we get: \frac{1}{2} < x - 1 < \frac{3}{2} \ \Rightarrow {\color{red}2 > \frac{1}{|x - 1|}} > \frac{2}{3}

    Can you finish off?
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  3. #3
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    Can you finish it please?

    I understand what you've done so far, but don't know what to do next. :P
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    It isn't that big a leap.

    You want to get |x - 2| < \text{Something}. Then we'll let \delta = \text{Something} and we're done.

    We have: 5\frac{{\color{red}1}|x-2|}{{\color{red}|x-1|}} < \epsilon and {\color{red}\frac{1}{|x-1|} < 2}.

    So we know that: 5\frac{|x-2|}{|x-1|} < \hdots ??? Think!
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  5. #5
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    ε / 10 = δ?
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    Quote Originally Posted by coldfire View Post
    or  \delta = \frac{2\epsilon}{5}?

    I'm so confused
    we have 5 \cdot \frac {|x - 2|}{|x - 1|} < \epsilon with \frac 1{|x - 1|} < 2

    thus, 5 \cdot \frac {|x - 2|}{|x - 1|} < 5 \cdot 2 |x - 2| = 10|x - 2| < \epsilon

    so ...
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  7. #7
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    so it's \delta = \frac{\epsilon}{10}?
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    Quote Originally Posted by coldfire View Post
    so it's \delta = \frac{\epsilon}{10}?
    yes
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  9. #9
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    ok thanks guys

    for the answer, do I write:

    \delta = \frac{\epsilon}{10} if \frac{1}{|x-1|} < 2?

    or if not, what do I put after if?
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    Quote Originally Posted by coldfire View Post
    ok thanks guys

    for the answer, do I write:

    \delta = \frac{\epsilon}{10} if \frac{1}{|x-1|} < 2?

    or if not, what do I put after if?
    no. remember, you are trying to prove that 7 is thew limit. restate the definition of what the limit is and say something along the lines of, "choose \delta = \frac {\epsilon}{10}, then blah bla blah is less than \epsilon, as desired"
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  11. #11
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    Actually it's \delta = \text{min} \left( \frac{1}{2}, \frac{\epsilon}{10}\right) because you have 2 inequalities to be satisfied: |x - 2| < \frac{1}{2} and |x-2| < \frac{\epsilon}{2}.

    The lower of the two will satisfy both, hence our choice of \delta
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by o_O View Post
    Actually it's \delta = \text{min} \left( \frac{1}{2}, \frac{\epsilon}{10}\right) because you have 2 inequalities to be satisfied: |x - 2| < \frac{1}{2} and |x-2| < \frac{\epsilon}{2}.

    The lower of the two will satisfy both, hence our choice of \delta
    yep, i forgot that. we were concentrating so hard on solving that inequality
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  13. #13
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    Quote Originally Posted by o_O View Post
    \lim_{x \to 2} \frac{2x + 3}{x-1} = 7

    (Normally we use \delta \leq 1 but we run into a bit of a problem so we choose a different value).
    Sorry to bump this problem but o_O, could you explain why you chose 1/2 instead of 1, I used 1 and I thought I had achieved the same result. I read your explanation but in which cases should I consider a value other than 1? I'm having a hard time trying to grasp this
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  14. #14
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    The thing is you don't have to work with \delta \leq 1. Generally, this will work but sometimes it just messes up the algebra so just pick another small value for \delta.

    If we used \delta \leq 1, we would have: |x - 2| < \delta = 1 \ \Rightarrow \ -1 \ < \ x-2 \ < \ 1

    Adding +1 to both sides we get: 0 \ < \  |x - 1| \ < \ 2

    Now the point of doing all this is to get: \frac{1}{|x-1|} < a where a is some constant.

    As we can see, if we tried to take the reciprical of all three sides (and thereby switching the inequality around), we would have:
    \frac{1}{2} < \frac{1}{|x-1|} < "\frac{1}{0}"

    which means we don't have an upper bound for \frac{1}{|x-1|}.

    So I picked another value for \delta that won't cause this problem.
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    Thumbs up

    Thank you for your explanation o_0!
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