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

Question

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

Charles Arocha · Accepted Answer

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

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 #

Isaiah Allen · Answer

undefined To find the limit of (3+2√u)/(4-√u) as u approaches infinity, we can use the concept of rationalizing the denominator. By multiplying both the numerator and denominator by the conjugate of the denominator, which is (4+√u), we can simplify the expression.

After multiplying, we get:
[(3+2√u)(4+√u)] / [(4-√u)(4+√u)]

Expanding the numerator and denominator, we have:
(12 + 7√u + 2u) / (16 - u)

As u approaches infinity, the terms involving √u and u will dominate the expression. Since the highest power of u is in the numerator, we can ignore the denominator.

Therefore, the limit of (3+2√u)/(4-√u) as u approaches infinity is ∞.

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

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