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Thread: Proving this $\epsilon$ $\delta$ proof

  1. #1
    Senior Member x3bnm's Avatar
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    Proving this $\epsilon$ $\delta$ proof

    $\text{Proving the $\epsilon$ $\delta$ proof}$

    Is it possible to check this work of mine?


    $\text{Show that $\displaystyle \lim_{x \to -1} \frac{3x +2}{x+2} = -1$}$


    $\text{We want $|\frac{3x +2}{x+2} +1| < \epsilon$ when $|x+1| <\delta$}$





    $\text{We will try to solve the inequality $|\frac{3x +2}{x+2} +1| < \epsilon$ for $|x+1|$}$



    $\displaystyle |\frac{3x +2}{x+2} +1| < \epsilon$



    $\displaystyle \frac{|1 + x|}{|x+2|}<\frac{\epsilon}{4}$



    $\displaystyle \frac{|1+x|}{|x+2|}<\frac{\epsilon}{4}$



    $\displaystyle \text{Let } \frac{1}{|x+2|} \le M$



    Because the you cannot isolate $|x+1|$ completely that has independent variable $\epsilon$
    Let $\displaystyle \delta = \frac{-3}{4}$



    $\displaystyle {|x + 1|} > \frac{-3}{4}$




    $\displaystyle \frac{7}{4}<\frac{1}{x + 2} <4$




    \begin{align*}|\frac{3x +2}{x+2} +1| &= \frac{4|x+1|}{|x+2|}\\&< 4|x+1| \text{.....[$|x +1|< \delta $ and $\delta \le \frac{-3}{4}$]}\\&< 4\frac{\epsilon}{4}\text{.....[$|x + 1|< \delta $ and $\delta \le \frac{\epsilon}{4}$]}\\&=\epsilon\end{align*}
    Last edited by x3bnm; Aug 13th 2018 at 06:57 AM.
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  2. #2
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    Re: Proving this $\epsilon$ $\delta$ proof

    You need $0<\delta$ so a negative number will not work.

    Let's try bounding the denominator. You have $|x+1|<\delta$, so $-\delta < x+1 < \delta$. Adding 1, we get: $1-\delta < x+2 < 1+\delta$. So, as $\delta \to 1$, we pick up values of $x$ that make that fraction infinite. However, we can limit the upper bound of $\delta$ to whatever we want. Let's assume we place an upper bound on $\delta$ of $\dfrac{1}{2}$. We know that $0 < \delta < \dfrac{1}{2}$. Then, we have:

    $$\dfrac{1}{2} < x+2 < \dfrac{3}{2}$$

    Plugging in, we get:

    $$4\left| \dfrac{x+1}{x+2} \right| < 4\left|\dfrac{x+1}{\dfrac{1}{2}} \right| = 8|x+1| < 8\delta$$

    So, if $\delta = \text{min}\left( \dfrac{\epsilon}{8}, \dfrac{1}{2}\right)$, then you get:

    $$8|x+1|<8\delta < \epsilon$$
    Last edited by SlipEternal; Aug 13th 2018 at 07:21 AM.
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  3. #3
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    Re: Proving this $\epsilon$ $\delta$ proof

    Quote Originally Posted by x3bnm View Post
    $\text{Proving the $\epsilon$ $\delta$ proof}$
    $\text{Show that $\displaystyle \lim_{x \to -1} \frac{3x +2}{x+2} = -1$}$
    $\text{We want $|\frac{3x +2}{x+2} +1| < \epsilon$ when $|x+1| <\delta$}$
    $\displaystyle |\frac{3x +2}{x+2} +1| < \epsilon$
    First understand that this in no way is negative criticism of the previous answer.
    But this is just my way of teaching this. You have found that we need that $4\left| {\frac{{x + 1}}{{x + 2}}} \right| < \varepsilon $ provided that $|x+1|<\delta$
    Now in most cases I ask students to consider $|x+1|<1$ but in this case it will not work.
    So lets say:
    $ \begin{align*} \left| {x + 1} \right| &< \frac{1}{3} \\- \frac{1}{3} < x + 1 &< \frac{1}{3} \\ \frac{2}{3}< x + 2 &< \frac{4}{3} \\\frac{1}{{x + 2}}& < \frac{3}{2} \end{align*}$

    If $\delta = \min \left\{ {\frac{1}{3},\frac{\varepsilon }{8}} \right\}$ and $|x+1|<\delta$ then
    $4\left| {\frac{{x + 1}}{{x + 2}}} \right| < 4 \cdot \frac{3}{2} \cdot \frac{\varepsilon }{8} = \frac{{3\varepsilon }}{4} < \varepsilon$
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