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Math Help - Limit Problem

  1. #1
    Junior Member
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    Limit Problem

    Could someone show the steps in working out this limit:

    limit (k^2*sqrt(9k^2-4k+1))/(3k^3+k-7) as k goes from 3 to infinity

    thanks
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Sam1111 View Post
    Could someone show the steps in working out this limit:

    limit (k^2*sqrt(9k^2-4k+1))/(3k^3+k-7) as k goes from 3 to infinity

    thanks
    \mathop {\lim }\limits_{x \to \infty } \frac{{{k^2}\sqrt {9{k^2} - 4k + 1} }}{{3{k^3} + k - 7}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{k^3}\sqrt {9 - \frac{4}<br />
{k} + \frac{1}<br />
{{{k^2}}}} }}<br />
{{{k^3}\left( {3 + \frac{1}<br />
{{{k^2}}} - \frac{7}<br />
{{{k^3}}}} \right)}} =

    = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {9 - \frac{4}<br />
{k} + \frac{1}<br />
{{{k^2}}}} }}<br />
{{3 + \frac{1}<br />
{{{k^2}}} - \frac{7}<br />
{{{k^3}}}}} = \frac{{\sqrt {9 - 0 + 0} }}<br />
{{3 + 0 - 0}} = \frac{3}<br />
{3} = 1.
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  3. #3
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    Cheers, I'm not sure why I didn't see that
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