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Math Help - sequence2!

  1. #1
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    sequence2!

    evaluate  (x-\sqrt{x^2-a^2}) as x tends to infinity

    pls how do i go about the problem above, thanks.
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  2. #2
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    Re: sequence2!

    Quote Originally Posted by lawochekel View Post
    evaluate  (x-\sqrt{x^2-a^2}) as x tends to infinity

    pls how do i go about the problem above, thanks.
    Multiplying top and bottom by the conjugate might be a good idea...

    \displaystyle \begin{align*} x - \sqrt{ x^2 - a^2 } &= \left( x - \sqrt{ x^2 - a^2 } \right) \left( \frac{x + \sqrt{x^2 - a^2} }{x + \sqrt{x^2 - a^2} } \right) \\ &= \frac{x^2 - \left(x^2 - a^2\right)}{x + \sqrt{x^2-a^2}} \\ &= \frac{a^2}{x + \sqrt{x^2 - a^2}}\end{align*}

    I'm sure you can evaluate this limit as \displaystyle \begin{align*} x \to \infty \end{align*}.
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