Calculate the limit

[flaren5]

lim (√x² + 2x - √x² - 2x )
x→∞


I seem to be encountering difficulties with the infinity limit...I'm aware that it's relatively easy, I just seem to be making it harder than it is. If anyone has a way of explaining it in "simple" terms, it would be greatly appreciated.

Thank you.


*Note* (x² + 2x) and (x² - 2x) are both under a square root, not just the x².

 

[pka]

lim (√x² + 2x - √x² - 2x )x→∞

\(\displaystyle \sqrt{x^2+2x}-\sqrt{x^2-2x}=\dfrac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}} \)

 

[DrPhil]

lim (√x² + 2x - √x² - 2x )
x→∞


I seem to be encountering difficulties with the infinity limit...I'm aware that it's relatively easy, I just seem to be making it harder than it is. If anyone has a way of explaining it in "simple" terms, it would be greatly appreciated.

Thank you.


*Note* (x² + 2x) and (x² - 2x) are both under a square root, not just the x².

There is obviously more than one way to do this one, but my favorite is to expand the square roots as power series, \(\displaystyle \sqrt{1 + \delta} \approx 1 + \delta /2\). Then

\(\displaystyle \displaystyle \sqrt{x^2 + 2 x} - \sqrt{x^2 - 2 x}\ =\ x\ \left[ \sqrt{1 + \dfrac{2}{x}} - \sqrt{1 - \dfrac{2}{x}} \right]\)

.......................................\(\displaystyle \displaystyle \approx \ x\ \left[ \left(1 + \dfrac{1}{x} \right) - \left(1 - \dfrac{1}{x} \right) \right] = \cdot \cdot \cdot\)

 

[flaren5]

Thank you for an example of both ways to do this limit.
I did figure it out the way DrPhil showed how to do it....I'm still a little rusty on how to expand the square root with the power series, but i do find it easier when I remember the rules.