Question

How do you find `lim (3+u^(-1/2)+u^-1)/(2+4u^(-1/2))` as `u->0^+` using l'Hospital's Rule?

Answer 1

l"Hospital's Rule is neither needed nor helpful for this limit.

Explanation:

We do get the indeterminate form `oo/oo`, so we can try l'Hospital's rile, but we'll get and continue to get negative powers of `u` in both the numerator and the denominator.
Therefore, we will continue to get the indeterminate `oo/oo`

Multiply by `u/u`, to get

`(3+u^(-1/2)+u^-1)/(2+4u^(-1/2)) = (3u+u^(1/2)+1)/(2u+4u^(1/2))`

We no longer get an indeterminate form, so we cannot use (and have no need for) l'Hospital's Rule.

`lim_(xrarr0^+)(3u+u^(1/2)+1)/(2u+4u^(1/2)) = oo`

(The numerator is going to a positive limit and the denominator is going to `0` through positive value.)