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Math Help - Finding limits w/definition not shortcuts

  1. #1
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    Question Finding limits w/definition not shortcuts

    My question is how to get the limit of lim ((3n+1)/(n+2))=3.

    I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.


    However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.

    Any suggestions?
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  2. #2
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    Re: Finding limits w/definition not shortcuts

    Quote Originally Posted by CountingPenguins View Post
    My question is how to get the limit of lim ((3n+1)/(n+2))=3.

    I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.
    I'm sort of confused by your post. So are you suppose to prove that \lim\frac{3n+1}{n+2}=3?

    If yes, then start with the inequality |\frac{3n+1}{n+2}-3|=|\frac{3n+1}{n+2}-\frac{3(n+2)}{n+2}|<|\frac{-5}{n}|.
    Last edited by Joanna; June 26th 2011 at 02:11 AM. Reason: Latex is being weird, fixed an inequality
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finding limits w/definition not shortcuts

    Quote Originally Posted by CountingPenguins View Post
    Any suggestions?
    Some more details: \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2 now, for every \epsilon>0 choose N>\dfrac{5}{ \epsilon}-2 .
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  4. #4
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    Re: Finding limits w/definition not shortcuts

    [QUOTE=FernandoRevilla;662680]Some more details: \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2


    Thank you for your response, but can you tell me how you got from \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2 it isn't obvious to me. Those are the details I am missing.

    Thanks,
    CP
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finding limits w/definition not shortcuts

    \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \left |\dfrac{3n+1-3n-6}{n+2}\right |< \epsilon\Leftrightarrow

    \left |\dfrac{-5}{n+2}\right |< \epsilon\Leftrightarrow \dfrac{5}{n+2}< \epsilon\Leftrightarrow\ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2
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  6. #6
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    Re: Finding limits w/definition not shortcuts

    Quote Originally Posted by CountingPenguins View Post
    My question is how to get the limit of lim ((3n+1)/(n+2))=3.

    I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.


    However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.

    Any suggestions?
    To prove \displaystyle \lim_{x \to \infty}f(x) = L, you need to show that \displaystyle x > N \implies |f(x) - L| < \epsilon for \displaystyle \epsilon > 0. So in this case, to prove \displaystyle \lim_{n \to \infty}\frac{3n + 1}{n + 2} = 3, you need to show \displaystyle n > N \implies \left|\frac{3n + 1}{n + 2} - 3\right| < \epsilon.


    \displaystyle \begin{align*} \left|\frac{3n + 1}{n + 2} - 3\right| &< \epsilon \\ \left|\frac{3n + 1 - 3(n + 2)}{n + 2}\right| &< \epsilon \\ \left|-\frac{5}{n+2}\right| &< \epsilon \\ \frac{5}{|n + 2|} &< \epsilon \\ \frac{|n + 2|}{5} &> \frac{1}{\epsilon} \\ |n + 2| &> \frac{5}{\epsilon} \\ n + 2 &> \frac{5}{\epsilon}\textrm{ since }\epsilon > 0\textrm{ and }n > 0 \\ n &> \frac{5}{\epsilon} - 2\end{align*}

    So by letting \displaystyle N = \frac{5}{\epsilon} - 2, the proof will follow.
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  7. #7
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    Re: Finding limits w/definition not shortcuts

    Thank you very much for filling in all the blanks for me. I don't want the answer so much as how and why it works and you showed me how. Thanks.
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