# How do you find the limit of (t^2-4)/(t^3-8) as t approaches 2?

Nov 17, 2016

$\lim f {\left(t\right)}_{t = 2} = \frac{1}{3}$

#### Explanation:

$f \left(t\right) = \frac{{t}^{2} - 4}{{t}^{3} - 8}$ for $t = 2$ gives $\frac{0}{0}$

is possible to use L'Hôpital's rule

$\lim f {\left(t\right)}_{t = 2} = \lim \frac{2 t}{3 {t}^{2}} = \lim \frac{2}{3 t} = \frac{1}{3}$

Nov 17, 2016

An algebraic solution:

${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = {\lim}_{t \rightarrow 2} \frac{{t}^{2} - {2}^{2}}{{t}^{3} - {2}^{3}}$
${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = {\lim}_{t \rightarrow 2} \frac{\left(t + 2\right) \left(t - 2\right)}{\left(t - 2\right) \left({t}^{2} + 2 t + 4\right)}$
${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = {\lim}_{t \rightarrow 2} \frac{\left(t + 2\right)}{\left({t}^{2} + 2 t + 4\right)}$
${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = \frac{2 + 2}{4 + 4 + 4}$
${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = \frac{4}{12}$
${\lim}_{t \rightarrow 2} \frac{{t}^{2} - 4}{{t}^{3} - 8} = \frac{1}{3}$