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Math Help - Limits at Negative Infinity

  1. #1
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    Limits at Negative Infinity

    Lim {x -> -∞} (√x^2 - 6x +3) / 2x + x

    Please help me evaluate the limit
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  2. #2
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    Quote Originally Posted by valeriegriff View Post
    Lim {x -> -∞} (√x^2 - 6x +3) / 2x + x

    Please help me evaluate the limit
    Do you mean this \mathop {\lim }\limits_{x \to  - \infty } \left[ {\frac{{\sqrt {{x^2} - 6x + 3} }}{{2x}} + x} \right]

    or \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - 6x + 3} }}{{2x + 1}} ??
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  3. #3
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    Re:

    Sorry, I meant


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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by valeriegriff View Post
    Sorry, I meant


    \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - 6x + 3} }}{{2x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {1 - \frac{6}<br />
{x} + \frac{3}<br />
{{{x^2}}}} }}<br />
{{x\left( {2 + \frac{1}<br />
{x}} \right)}} =

    =  - \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {1 - \frac{6}<br />
{x} + \frac{3}<br />
{{{x^2}}}} }}<br />
{{2 + \frac{1}<br />
{x}}} =  - \frac{{\sqrt {1 - 0 + 0} }}<br />
{{2 - 0}} =  - \frac{1}{2}.
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