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29 June, 05:42

Use algebra to simplify the expression before evaluating the limit. In particular, factor the highest power of n from the numerator and denominator, then cancel as many factors of n as possible. If the sequence does not converge, enter diverges in the final answer box. limn→[infinity]2-7n+9n28n2+2n-6

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  1. 29 June, 08:46
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    a) {2/n²-7/n+9}/{8+2/n-6/n²}

    b) 9/8

    c) The sequence converges

    Step-by-step explanation:

    Given the limit of the function

    limn→[infinity]2-7n+9n²/8n²+2n-6

    To simplify the function given, we will have to factor out the highest power of n which is n² from the numerator and the denominator. The function will then become;

    2-7n+9n²/8n²+2n-6

    = n²{2/n²-7/n+9}/n²{8+2/n-6/n²}

    The n² at the numerator will then cancel out the n² at the denominator to have resulting simplified equation as;

    {2/n²-7/n+9}/{8+2/n-6/n²}

    Evaluating the limit of the resulting equation will give;

    limn→[infinity] {2/n²-7/n+9}/{8+2/n-6/n²}

    Note that limn→[infinity] a/n = 0 where a is any constant.

    Therefore;

    limn→[infinity] {2/n²-7/n+9}/{8+2/n-6/n²}

    = (0-0+9) / (8+0-0)

    = 9/8

    Since the limit of the sequence gives a finite value which is 9/8, thus the sequence in question is a convergent sequence.

    The limit of a sequence only diverges if the limit of such sequence is an infinite value.
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