# How do you find lim sqrt(u^2-3u+2)-sqrt(u^2+1) as u->oo?

Apr 12, 2017

$- \frac{3}{2}$

#### Explanation:

Note that:

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

$= \frac{\left({u}^{2} - 3 u + 2\right) - \left({u}^{2} + 1\right)}{\sqrt{{u}^{2} - 3 u + 2} + \sqrt{{u}^{2} + 1}}$

$= \frac{- 3 u + 1}{\sqrt{{u}^{2} - 3 u + 2} + \sqrt{{u}^{2} + 1}}$

Factoring out the terms with the largest degree:

$= \frac{u \left(- 3 + \frac{1}{u}\right)}{\sqrt{{u}^{2} \left(1 - \frac{3}{u} + \frac{2}{u} ^ 2\right)} + \sqrt{{u}^{2} \left(1 + \frac{1}{u} ^ 2\right)}}$

$= \frac{u \left(- 3 + \frac{1}{u}\right)}{\left\mid u \right\mid \left(\sqrt{1 - \frac{3}{u} + \frac{2}{u} ^ 2} + \sqrt{1 + \frac{1}{u} ^ 2}\right)}$

Also note that:

$\left\mid u \right\mid = \left\{\begin{matrix}u & \text{ & " & u>0 \\ -u & " & } & u < 0\end{matrix}\right.$

Since we're concerned with positive infinity, we say that $\left\mid u \right\mid = u$ in this case. The $u$ in the numerator and denominator then cancel.

$= \frac{- 3 + \frac{1}{u}}{\sqrt{1 - \frac{3}{u} + \frac{2}{u} ^ 2} + \sqrt{1 + \frac{1}{u} ^ 2}}$

So then:

${\lim}_{u \rightarrow \infty} \sqrt{{u}^{2} - 3 u + 2} - \sqrt{{u}^{2} + 1} = {\lim}_{u \rightarrow \infty} \frac{- 3 + \frac{1}{u}}{\sqrt{1 - \frac{3}{u} + \frac{2}{u} ^ 2} + \sqrt{1 + \frac{1}{u} ^ 2}}$

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

$= - \frac{3}{2}$