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#