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Math Help - 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 \delta > 0 that fulfills: if |x+1| < \delta, then \left| \frac{x}{2x+3} + 1 \right| < \frac{1}{100}

    Answer: \left| \frac{x}{2x+3} + 1 \right| = \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 x = -\frac{3}{2} so we can require 0 \leq \delta \leq \frac{1}{5} (so the x can stay away from -\frac{3}{2})

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

    So now we get

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

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

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

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

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

    Thanks!

    **edit thanks for your advice!
    Last edited by jayshizwiz; October 28th 2010 at 05:44 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) 0 \leq \delta \leq \frac{1}{5} but I guess since my final result was (2)  \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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