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

Question

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

Scarlett Allison · Accepted Answer

#-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#

Aurora Alvarado · Answer

undefined To find the limit of the expression lim sqrt(u^2-3u+2)-sqrt(u^2+1) as u approaches infinity, we can simplify the expression by rationalizing the denominator. By multiplying the expression by the conjugate of the denominator, we get:

lim 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))

Simplifying further, we have:

lim (u^2-3u+2) - (u^2+1) / (sqrt(u^2-3u+2)+sqrt(u^2+1))

This simplifies to:

lim -3u+1 / (sqrt(u^2-3u+2)+sqrt(u^2+1))

As u approaches infinity, the terms involving u^2 become negligible compared to the -3u term. Therefore, the limit becomes:

lim -3u / (sqrt(u^2)+sqrt(u^2))

Simplifying further, we have:

lim -3u / (u+u)

This simplifies to:

lim -3/2

Therefore, the limit of sqrt(u^2-3u+2)-sqrt(u^2+1) as u approaches infinity is -3/2.

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

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