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

## ${\lim}_{\text{x->oo}}$ $\sqrt{{x}^{2} - 4 x + 3} - \sqrt{{x}^{2} + 2 x}$

Oct 2, 2016

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

#### Explanation:

$\frac{\left(\sqrt{{x}^{2} - 4 x + 3} - \sqrt{{x}^{2} + 2 x}\right)}{1} \cdot \frac{\left(\sqrt{{x}^{2} - 4 x + 3} + \sqrt{{x}^{2} + 2 x}\right)}{\left(\sqrt{{x}^{2} - 4 x + 3} + \sqrt{{x}^{2} + 2 x}\right)}$

$= \frac{\left({x}^{2} - 4 x + 3\right) - \left({x}^{2} + 2 x\right)}{\sqrt{{x}^{2} - 4 x + 3} + \sqrt{{x}^{2} + 2 x}}$

$= \frac{- 6 x + 3}{\sqrt{{x}^{2} - 4 x + 3} + \sqrt{{x}^{2} + 2 x}}$

For $x \ne 0$, we can factor ${x}^{2}$ under the radicals,

$= \frac{- 6 x + 3}{\sqrt{{x}^{2}} \sqrt{1 - \frac{4}{x} + \frac{3}{x} ^ 2} + \sqrt{{x}^{2}} \sqrt{1 + \frac{2}{x}}}$

Now, since $\sqrt{{x}^{2}} = \left\mid x \right\mid$ abd we are only interested in $x > 0$ as $x \rightarrow \infty$, we have

$= \frac{- 6 x + 3}{x \sqrt{1 - \frac{4}{x} + \frac{3}{x} ^ 2} + x \sqrt{1 + \frac{2}{x}}}$ (for $x > 0$)

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

$= \frac{\cancel{x} \left(- 6 + \frac{3}{x}\right)}{\cancel{x} \left(\sqrt{1 - \frac{4}{x} + \frac{3}{x} ^ 2} + \sqrt{1 + \frac{2}{x}}\right)}$ (for $x > 0$).

Taking the limit as $x \rightarrow \infty$ gets us

$\frac{- 6 + 0}{\sqrt{1 - 0 + 0} + \sqrt{1 + 0}} = \frac{- 6}{2} = - 3$