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Thread: Beginners calculus type of question (...I think)

  1. #1
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    Beginners calculus type of question (...I think)

    Hi. I answered this question but I have no idea if I did it correctly...

    Question: Find a $\displaystyle \delta > 0$ that fulfills: if $\displaystyle |x+1| < \delta$, then $\displaystyle \left| \frac{x}{2x+3} + 1 \right| < \frac{1}{100}$

    Answer: $\displaystyle \left| \frac{x}{2x+3} + 1 \right|$ = $\displaystyle \left| \frac{3x+3}{2x+3} \right| = \frac{3|x+1|}{|2x+3|}$

    That part was easy, now is the part I don't know if I did logically/correctly...

    The function is not defined at $\displaystyle x = -\frac{3}{2}$ so we can require $\displaystyle 0 \leq \delta \leq \frac{1}{5}$ (so the x can stay away from $\displaystyle -\frac{3}{2}$)

    so that $\displaystyle |x+1| < \frac{1}{5}$ $\displaystyle \rightarrow$ $\displaystyle - \frac{1}{5} \leq x+1 < \frac{1}{5}$ $\displaystyle \rightarrow$ $\displaystyle 0 < \frac{3}{5}< 2x+3 < \frac{7}{5}$$\displaystyle \rightarrow$$\displaystyle |2x+3|>\frac{3}{5}$

    So now we get

    $\displaystyle \frac{3|x+1|}{|2x+3|} \leq \frac{3|x+1|}{\frac{3}{5}} = 5|x+1|$

    and since $\displaystyle |x+1| < \delta$, $\displaystyle 5|x+1|<5\delta$

    Therefore if we require $\displaystyle 5\delta < \frac{1}{100}$, we will for sure get for $\displaystyle |x+1|< \delta$

    $\displaystyle 5|x+1| < 5\delta < \frac{1}{100}$

    and so $\displaystyle 0< \delta < \frac{1}{500}$

    Thanks!

    **edit thanks for your advice!
    Last edited by jayshizwiz; Oct 28th 2010 at 04:44 AM.
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  2. #2
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    You were doing great until the very end. This line:
    5|x+1| < \delta < \frac{1}{100}, should be
    5|x+1| < 5\delta < \frac{1}{100}. You corrected it in the next line, though.
    Also, you found that delta < 1/500. Great! Stop right there. You're done. delta < 1/5 does NOT satisfy the requirement.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    You were doing great until the very end. This line:
    5|x+1| < \delta < \frac{1}{100}, should be
    5|x+1| < 5\delta < \frac{1}{100}. You corrected it in the next line, though.
    Thanks! That was a typo. I'll edit it.

    Also, you found that delta < 1/500. Great! Stop right there. You're done. delta < 1/5 does NOT satisfy the requirement
    Thanks again. I just got confused since my initial statement was (1)$\displaystyle 0 \leq \delta \leq \frac{1}{5}$ but I guess since my final result was (2)$\displaystyle \delta < \frac{1}{500}$, as long as I choose a number smaller than (2), it also satisfies (1)....(as long as it's bigger than 0...)
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  4. #4
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    Looks like you're good to go, then. Yeah, you have to take the more restricted condition in cases like this.
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