evaluate $\displaystyle (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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- May 1st 2012, 04:22 AMlawochekelsequence2!
evaluate $\displaystyle (x-\sqrt{x^2-a^2}) $ as x tends to infinity

pls how do i go about the problem above, thanks. - May 1st 2012, 04:27 AMProve ItRe: sequence2!
Multiplying top and bottom by the conjugate might be a good idea...

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

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