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Math Help - Find the limit.

  1. #1
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    Find the limit.

    Find the limit.

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  2. #2
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    Quote Originally Posted by plstevens View Post
    Find the limit.

    \sqrt{25x^2 + x} - 5x = \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + 5x - 25x^2}{\sqrt{25x^2 + x} + 5x}

    = \frac{5x}{\sqrt{25x^2 + x} + 5x} = \frac{5}{\sqrt{25 + \frac{1}{x}} + 5} .

    Take the limit as x \rightarrow \infty of that.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    \sqrt{25x^2 + x} - 5x = \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + {\color{red} 5x} - 25x^2}{\sqrt{25x^2 + x} + 5x}
    ...[/tex]

    Take the limit as x \rightarrow \infty of that.
    you have a small typo mr fantastic. i was going to ask plstevens to find it, but i said never mind. the term highlighted should be x
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  4. #4
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    i don't understand how u got 5x, when u multiple the top out u get 25x^2+x+5x? cause i still don't understand how u got what u got
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  5. #5
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    Quote Originally Posted by plstevens View Post
    i don't understand how u got 5x, when u multiple the top out u get 25x^2+x+5x? cause i still don't understand how u got what u got
    To simplify \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)},

    recall: (A - B)(A + B) = A^2 - B^2.

    Let A = \sqrt{25x^2 + x} and B = 5x.

    Then A^2 - B^2 = 25x^2 + x - (5x)^2 = 25x^2 + x - 25x^2 = x.

    Therefore <br />
\frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + x - 25x^2}{\sqrt{25x^2 + x} + 5x} = \frac{x}{\sqrt{25x^2 + x} + 5x}.<br />



    Typo noted - thanks for that get, Jhevon.
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