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Thread: 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
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    $\displaystyle \lim_{x \to 2} \frac{2x + 3}{x-1} = 7$

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

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

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

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

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

    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 $\displaystyle |x - 2| < \text{Something}$. Then we'll let $\displaystyle \delta = \text{Something}$ and we're done.

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

    So we know that: $\displaystyle 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 $\displaystyle \delta = \frac{2\epsilon}{5}?$

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

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

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

    for the answer, do I write:

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

    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:

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

    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 $\displaystyle \delta = \frac {\epsilon}{10}$, then blah bla blah is less than $\displaystyle \epsilon$, as desired"
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  11. #11
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    Actually it's $\displaystyle \delta = \text{min} \left( \frac{1}{2}, \frac{\epsilon}{10}\right)$ because you have 2 inequalities to be satisfied: $\displaystyle |x - 2| < \frac{1}{2}$ and $\displaystyle |x-2| < \frac{\epsilon}{2}$.

    The lower of the two will satisfy both, hence our choice of $\displaystyle \delta$
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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 $\displaystyle \delta = \text{min} \left( \frac{1}{2}, \frac{\epsilon}{10}\right)$ because you have 2 inequalities to be satisfied: $\displaystyle |x - 2| < \frac{1}{2}$ and $\displaystyle |x-2| < \frac{\epsilon}{2}$.

    The lower of the two will satisfy both, hence our choice of $\displaystyle \delta$
    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
    $\displaystyle \lim_{x \to 2} \frac{2x + 3}{x-1} = 7$

    (Normally we use $\displaystyle \delta \leq 1$ 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 $\displaystyle \delta \leq 1$. Generally, this will work but sometimes it just messes up the algebra so just pick another small value for $\displaystyle \delta$.

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

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

    Now the point of doing all this is to get: $\displaystyle \frac{1}{|x-1|} < a$ where $\displaystyle a$ 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:
    $\displaystyle \frac{1}{2} < \frac{1}{|x-1|} < "\frac{1}{0}"$

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

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

    Thank you for your explanation o_0!
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