# How do you find lim (t+1/t)((4-t)^(3/2)-8) as t->0 using l'Hospital's Rule?

Aug 14, 2017

#### Explanation:

$\left(t + \frac{1}{t}\right) \left({\left(4 - t\right)}^{\frac{3}{2}} - 8\right)$ as $t \rightarrow 0$ has form $\infty \cdot 0$.

We will rewrite it as ((4-t)^(3/2)-8)/(1/(t+1/t) so we have an expression whose limit has form $\frac{0}{0}$ and we can use l'Hospital's Rule.

Note that $t + \frac{1}{t} = \frac{{t}^{2} + 1}{t}$ so we want

lim_(trarro)(t+1/t)((4-t)^(3/2)-8) = lim+(trarr0)((4-t)^(3/2)-8)/(1/(t+1/t)

$= {\lim}_{t \rightarrow 0} \frac{{\left(4 - t\right)}^{\frac{3}{2}} - 8}{\frac{t}{{t}^{2} + 1}}$

$= {\lim}_{t \rightarrow 0} \frac{- \left(\frac{3}{2}\right) {\left(4 - t\right)}^{\frac{1}{2}}}{\frac{1 \left({t}^{2} + 1\right) - t \left(2 t\right)}{{t}^{2} + 1} ^ 2}$

$= {\lim}_{t \rightarrow 0} \frac{- \left(\frac{3}{2}\right) {\left(4 - t\right)}^{\frac{1}{2}}}{\frac{1 - {t}^{2}}{{t}^{2} + 1} ^ 2}$

$= \frac{\left(- \frac{3}{2}\right) \left(2\right)}{\frac{1}{1} ^ 2} = - 3$