# Math Help - sequence2!

1. ## sequence2!

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

pls how do i go about the problem above, thanks.

2. ## Re: sequence2!

Originally Posted by lawochekel
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*}.