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Thread: Limits at Infinity

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    Limits at Infinity

    Lim {x - > ∞ } (x - √x^2 + x)



    Evaluate the Limit
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by valeriegriff View Post
    Lim {x - > ∞ } (x - √x^2 + x)



    Evaluate the Limit
    Either I've missed something or the question is wrong:

    $\displaystyle x - \sqrt{x^2} + x = x-x+x = 2x-x = x$

    so the limit would be $\displaystyle \infty$
    Last edited by e^(i*pi); Oct 1st 2009 at 01:14 PM. Reason: ******* latex
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by valeriegriff View Post
    Lim {x - > ∞ } (x - √x^2 + x)


    Evaluate the Limit
    $\displaystyle \mathop {\lim }\limits_{x \to \infty } \left( {x - \sqrt {{x^2} + x} } \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - \sqrt {{x^2} + x} } \right)\left( {x + \sqrt {{x^2} + x} } \right)}}
    {{x + \sqrt {{x^2} + x} }} =$

    $\displaystyle = - \mathop {\lim }\limits_{x \to \infty } \frac{x}
    {{x + \sqrt {{x^2} + x} }} = - \mathop {\lim }\limits_{x \to \infty } \frac{1}
    {{1 + \sqrt {1 + \frac{1}
    {{{x^2}}}} }} = - \frac{1}
    {{1 + \sqrt {1 + 0} }} = - \frac{1}
    {2}.$
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