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Thread: limit problem

  1. #1
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    limit problem

    Prove $\displaystyle \displaystyle \lim_{x\to 1}\frac{x+2}{x^2+2}=1$ using $\displaystyle \: \epsilon, \: \delta$

    Attempt:

    $\displaystyle \: \epsilon >0$.
    We must find $\displaystyle \delta >0$ so that every $\displaystyle x$ that fulfills $\displaystyle 0<|x-1|< \delta \:\:\:$ fulfills $\displaystyle |\frac{x+2}{x^2+2} -1| < \epsilon$

    $\displaystyle |\frac{x+2}{x^2+2}-1| = |\frac{x-x^2}{x^2+2}| = \frac{|x||1-x|}{|x^2+2|} = \frac{|x||x-1|}{|x^2+2|}$

    I'm not sure about this part:

    $\displaystyle \frac{|x||x-1|}{|x^2+2|} \leq_{\downarrow} \frac{|x||x-1|}{|x|} = |x-1| < \epsilon$
    since $\displaystyle _{|x^2+2| \geq |x|}$

    and since
    $\displaystyle 0<|x-1|< \delta$,

    for every
    $\displaystyle 0 < \delta \leq \epsilon$ the required is fulfilled.


    Is this correct?
    Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    How about the same idea with the squeeze theorem?

    $\displaystyle 0<\left|{x+2\over x^2+2}-1\right|=\left|{x-x^2\over x^2+2}\right|={|x||1-x|\over x^2+2}< |1-x|\to 0$
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  3. #3
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    you can use the squeeze theorem, but the question specifically says use epsilon delta proof.

    But does it look right the way I did it?
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  4. #4
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    yes i almost sure that you are right .
    Last edited by imiviortal; Dec 2nd 2010 at 02:14 AM.
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  5. #5
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    No longer necessary.
    Last edited by HallsofIvy; Dec 2nd 2010 at 03:03 AM.
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  6. #6
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    yes i know ive edited my message one minute before youve posted hehe (3:14)
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  7. #7
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    I didn't see what you guys originally wrote but i'm going to assume it was funny.
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  8. #8
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    Not funny- imiviortal had a typo. I pointed it out but he had corrected it while I was typing so I removed my post.
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