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Math Help - Using epsilon-N definition to prove limits

  1. #1
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    Using epsilon-N definition to prove limits

    Hi,
    im stuck with a question on proving a limit of a sequence. Here it is:

    Use the epsilon-N definition of a limit to prove that:

    lim(n>infinity) n+5/n^2 + 2 = 0.

    For epsilon = 1/25 give N such that for all n>=N it follows that:

    |n+5/n^2 + 2| < epsilon.

    Not sure how to start really. Any help would be appreciated. Thanks
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  2. #2
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    Quote Originally Posted by leftfootwonder7 View Post
    Hi,
    im stuck with a question on proving a limit of a sequence. Here it is:

    Use the epsilon-N definition of a limit to prove that:

    lim(n>infinity) n+5/n^2 + 2 = 0.

    For epsilon = 1/25 give N such that for all n>=N it follows that:

    |n+5/n^2 + 2| < epsilon.

    Not sure how to start really. Any help would be appreciated. Thanks

    Try to add parentheses to make clear to others your expressions.

    Let \epsilon > 0 be arbitrary. We want to show that for all but a finite number of indexes n we have

    \mid\frac{n+5}{n^2+2}\mid<\epsilon

    As the expression between the absolute value is always positive we can drop the abs. value, so cross multiplying we get:

    n+5 <\epsilon n^2 + 2\epsilon \Longleftrightarrow \epsilon n^2-n+2\epsilon -5 >0

    Wel, now solve the quadratic inequality above and find your n!

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Try to add parentheses to make clear to others your expressions.

    Let \epsilon > 0 be arbitrary. We want to show that for all but a finite number of indexes n we have

    \mid\frac{n+5}{n^2+2}\mid<\epsilon

    As the expression between the absolute value is always positive we can drop the abs. value, so cross multiplying we get:

    n+5 <\epsilon n^2 + 2\epsilon \Longleftrightarrow \epsilon n^2-n+2\epsilon -5 >0

    Wel, now solve the quadratic inequality above and find your n!

    Tonio
    Hi Tonio, I don't understand why we couldn't just use the fact that:

     <br />
\frac{n+5}{n^2+2}< \frac{n+5}{n^2} = \frac{6}{n} < \epsilon<br />

    And then we can take N= \frac{6}{\epsilon}

    ??
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  4. #4
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    Quote Originally Posted by Roam View Post
    Hi Tonio, I don't understand why we couldn't just use the fact that:

     <br />
\frac{n+5}{n^2+2}< \frac{n+5}{n^2} = \frac{6}{n} < \epsilon<br />

    And then we can take N= \frac{6}{\epsilon}

    ??

    Who says we couldn't? Of course we could, it's just that you saw it and I didn't.

    Of course, you meant above \frac{5+n}{n^2} \leq \frac{6}{n}, and not equality.

    Well done...and you answered your own question.

    Tonio
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