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

Feb 19, 2017

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

#### Explanation:

We do get the indeterminate form $\frac{\infty}{\infty}$, 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 $\frac{\infty}{\infty}$

Multiply by $\frac{u}{u}$, to get

$\frac{3 + {u}^{- \frac{1}{2}} + {u}^{-} 1}{2 + 4 {u}^{- \frac{1}{2}}} = \frac{3 u + {u}^{\frac{1}{2}} + 1}{2 u + 4 {u}^{\frac{1}{2}}}$

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

${\lim}_{x \rightarrow {0}^{+}} \frac{3 u + {u}^{\frac{1}{2}} + 1}{2 u + 4 {u}^{\frac{1}{2}}} = \infty$

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