What is the limit of (t^(1/3)-2)/(t-8) as t approaches 8?

Sep 13, 2016

$\text{ The Limit=} \frac{1}{12}$.

Explanation:

There is a Standard Form of Limit :-

${\lim}_{x \rightarrow a} \frac{{x}^{n} - {a}^{n}}{x - a} = n \cdot {a}^{n - 1}$.

Using this Form, we have,

$\text{The Limit=} {\lim}_{t \rightarrow 8} \frac{{t}^{\frac{1}{3}} - 2}{t - 8}$

$= {\lim}_{t \rightarrow 8} \frac{{t}^{\frac{1}{3}} - {8}^{\frac{1}{3}}}{t - 8}$

$= \frac{1}{3} \cdot {8}^{\frac{1}{3} - 1}$

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

$\therefore \text{ The Limit=} \frac{1}{12}$, tallying with Claude D. 's Answer!

Sep 13, 2016

$\frac{1}{12}$.

Explanation:

We subst. $t = {\left(x + 2\right)}^{3} , \text{ with a note that, as } t \rightarrow 8 , x \rightarrow 0$.

Hence, the Reqd. Limit$= {\lim}_{x \rightarrow 0} \frac{{\left({\left(x + 2\right)}^{3}\right)}^{\frac{1}{3}} - 2}{{\left(x + 2\right)}^{3} - 8}$.

Here, we use, ${\left(a + b\right)}^{3} = {a}^{3} + {b}^{3} + 3 a b \left(a + b\right)$, to get,.

The Limit$= {\lim}_{x \rightarrow 0} \frac{\left(x + 2\right) - 2}{{x}^{3} + 8 + 3 \cdot x \cdot 2 \left(x + 2\right) - 8}$

$= {\lim}_{x \rightarrow 0} \frac{x}{{x}^{3} + 6 {x}^{2} + 12 x}$

$= {\lim}_{x \rightarrow 0} \frac{\cancel{x}}{\cancel{x} \left({x}^{2} + 6 x + 12\right)}$

$= {\lim}_{x \rightarrow 0} \frac{1}{{x}^{2} + 6 x + 12}$

$= \frac{1}{12}$, as before we had!

Enjoy Maths.!