Question

How do you find the limit as x approaches infinity of a more complex square root function?

`lim_"x->oo"` `sqrt(x^2-4x+3)-sqrt(x^2+2x)`

Answer 1

The initial form is indeterminate (`oo-oo`). One thing to try is writing it over `1` and 'rationalizing'. (Actually, we'll remove the square roots in the numerator.)

Explanation:

`((sqrt(x^2-4x+3)-sqrt(x^2+2x)))/1 * ((sqrt(x^2-4x+3)+sqrt(x^2+2x)))/((sqrt(x^2-4x+3)+sqrt(x^2+2x)))`

`= ((x^2-4x+3)-(x^2+2x))/(sqrt(x^2-4x+3)+sqrt(x^2+2x))`

`= (-6x+3)/(sqrt(x^2-4x+3)+sqrt(x^2+2x))`

For `x != 0`, we can factor `x^2` under the radicals,

`= (-6x+3)/(sqrt(x^2)sqrt(1-4/x+3/x^2)+sqrt(x^2)sqrt(1+2/x))`

Now, since `sqrt(x^2) = abs x` abd we are only interested in `x > 0` as `x rarroo`, we have

`= (-6x+3)/(xsqrt(1-4/x+3/x^2)+xsqrt(1+2/x))` (for `x > 0`)

Now factor out the `x`'a and reduce.

`= (cancel(x)(-6+3/x))/(cancel(x)(sqrt(1-4/x+3/x^2)+sqrt(1+2/x)))` (for `x > 0`).

Taking the limit as `xrarroo` gets us

`(-6+0)/(sqrt(1-0+0)+sqrt(1+0)) = (-6)/2 = -3`