How do you find #lim (3+2sqrtu)/(4-sqrtu)# as #u->oo#?

1 Answer
Mar 15, 2017

# lim_(u rarr oo)(3+2sqrt(u))/(4-sqrt(u)) = -2 #

Explanation:

We can manipulate the limit as follows;

# lim_(u rarr oo)(3+2sqrt(u))/(4-sqrt(u)) = lim_(u rarr oo)(3+2sqrt(u))/(4-sqrt(u)) * (1/sqrt(u))/(1/sqrt(u))#
# " " = lim_(u rarr oo)(3/sqrt(u)+2)/(4/sqrt(u)-1) #

# " " = (lim_(u rarr oo)(3/sqrt(u)+2)) / (lim_(u rarr oo)(4/sqrt(u)-1)) #

And as #u rarr oo => 1/sqrt(u) rarr 0 #; thus;

# lim_(u rarr oo)(3+2sqrt(u))/(4-sqrt(u)) = (0+2) / (0-1) #
# " " = 2/(-1) #
# " " = -2 #