Question

How do you find `lim (1+5/sqrtu)/(2+1/sqrtu)` as `u->0^+` using l'Hospital's Rule?

Answer 1

5

Answer 2

`lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u)) = 5`

Explanation:

We do not need to apply L'Hôpital's as this is a trivial limit to evaluate:

`lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u)) = lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u))*sqrt(u)/sqrt(u)`
`" " = lim_(u rarr 0^+) (sqrt(u)+5)/(2sqrt(u)+1)`
`" " = 5/1`

Note that the above limit is only valid for `u rarr 0^+` as:

`u rarr 0^+ => u > 0 => sqrt(u) in RR`

Whereas:

`u rarr 0^- => u < 0 => sqrt(u) in CC`