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Math Help - Evaluate the limit

  1. #1
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    Evaluate the limit

    lim (x^2+6x+1)^(1/2)-x
    x -> +inf

    I know the answer is 3, but I don't know how to find that without using a table.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Scary_Joe View Post
    lim (x^2+6x+1)^(1/2)-x
    x -> +inf

    I know the answer is 3, but I don't know how to find that without using a table.
    \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 6x + 1}  - x} \right) = \frac{{\left( {\sqrt {{x^2} + 6x + 1}  - x} \right)\left( {\sqrt {{x^2} + 6x + 1}  + x} \right)}}{{\left( {\sqrt {{x^2} + 6x + 1}  + x} \right)}} =

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

    = \mathop {\lim }\limits_{x \to  + \infty } \frac{{6 + \frac{1}<br />
{x}}}<br />
{{\sqrt {1 + \frac{6}<br />
{x} + \frac{1}<br />
{{{x^2}}}}  + 1}} = \frac{{6 + 0}}<br />
{{\sqrt {1 + 0 + 0}  + 1}} = \frac{6}<br />
{2} = 3.
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  3. #3
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    Wow, how did you get so good at evaluating limits? Because, when I took calculus I was never taught some of the techniques you used.
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