Question

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

Answer 1

`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`