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Math Help - Does the fraction simplify?

  1. #1
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    Does the fraction simplify?

    Hi all,

    I've been doing well all year, but now I've hit a wall. All the exercises prior to this question have had just one term of n, but now this has shown it's ugly self, and I am absolutely confused.

    The question regards sequences;

    For the sequence below, decide whether it converges and, if it does, state its limit.

    a(term n) = 6n^4 + 2n^3 (which is all over) 7n - 11n^4 (n=1,2,3,...)

    if the fraction simplifies so I can get either the numerator or the denominator a constant with no terms of n in, I should be OK then.

    Any help would be great.

    Shayne.
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  2. #2
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    Quote Originally Posted by Hydralisk View Post
    Hi all,

    I've been doing well all year, but now I've hit a wall. All the exercises prior to this question have had just one term of n, but now this has shown it's ugly self, and I am absolutely confused.

    The question regards sequences;

    For the sequence below, decide whether it converges and, if it does, state its limit.

    a(term n) = 6n^4 + 2n^3 (which is all over) 7n - 11n^4 (n=1,2,3,...)

    if the fraction simplifies so I can get either the numerator or the denominator a constant with no terms of n in, I should be OK then.

    Any help would be great.

    Shayne.
    For fractions like this, the technique is to divide top and bottom by the highest power of n in sight (in this case, n^4), and use the fact that negative powers of n go to zero as n\to\infty.
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  3. #3
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    That's helped, but I've still got terms of n in both the numerator and the denominator, so it's still "hard" for me.

    I've now got:

    (2/n)+ 6 (all over) (7/n^3)-11

    is there anything I can do to get rid of the (2/n) in the numerator maybe?

    Many thanks.
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  4. #4
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    Quote Originally Posted by Hydralisk View Post
    That's helped, but I've still got terms of n in both the numerator and the denominator, so it's still "hard" for me.

    I've now got:

    (2/n)+ 6 (all over) (7/n^3)-11

    is there anything I can do to get rid of the (2/n) in the numerator maybe?

    Many thanks.
    You just need to see that 2/n and 7/n^3 both get smaller and smaller as n increases. In the limit, they both tend to 0 (so you can ignore them altogether), and the fraction becomes 6/(–11) (which is the answer to the question).
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  5. #5
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    Hi,

    many thanks to you both, I did infact write down 6/-11 after a bit of time thinking it through, but having it confirmed is such a releif.

    Thanks again.
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  6. #6
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    Another way to think it through is.....

    As "n" approaches infinity, 6n^4 will dwarf 2n^3

    and -11n^4 will dwarf 7n

    so your fraction approaches

    \displaystyle-\frac{6n^4}{11n^4}=-\left(\frac{n^4}{n^4}\right)\frac{6}{11}=-\frac{6}{11}
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