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

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Answer 1 · 2017-04-12T21:50:16

$-3/2$

Note that:

$sqrt(u^2-3u+2)-sqrt(u^2+1)=(sqrt(u^2-3u+2)-sqrt(u^2+1))*(sqrt(u^2-3u+2)+sqrt(u^2+1))/(sqrt(u^2-3u+2)+sqrt(u^2+1))$

$=((u^2-3u+2)-(u^2+1))/(sqrt(u^2-3u+2)+sqrt(u^2+1))$

$=(-3u+1)/(sqrt(u^2-3u+2)+sqrt(u^2+1))$

Factoring out the terms with the largest degree:

$=(u(-3+1/u))/(sqrt(u^2(1-3/u+2/u^2))+sqrt(u^2(1+1/u^2)))$

$=(u(-3+1/u))/(absu(sqrt(1-3/u+2/u^2)+sqrt(1+1/u^2)))$

Also note that:

$absu={(u,",",u>0),(-u,",",u<0):}$

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

$=(-3+1/u)/(sqrt(1-3/u+2/u^2)+sqrt(1+1/u^2))$

So then:

$lim_(urarroo) sqrt(u^2-3u+2)-sqrt(u^2+1)=lim_(urarroo)(-3+1/u)/(sqrt(1-3/u+2/u^2)+sqrt(1+1/u^2))$

$=(-3+0)/(sqrt(1-0+0)+sqrt(1+0))$

$=-3/2$

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