# Evaluate the limit by using a change of variable?

Feb 8, 2017

Let u = ${\left(x + 8\right)}^{\frac{1}{3}}$

#### Explanation:

Then ${u}^{3} = x + 8$ and $x = {u}^{3} - 8$

As x approaches the value 0, u approaches the value 2. The given limit becomes

${\lim}_{x \to 0} \frac{{\left(x + 8\right)}^{\frac{1}{3}} - 2}{x} = {\lim}_{u \to 2} \frac{u - 2}{{u}^{3} - 8}$

$\frac{u - 2}{\left(u - 2\right) \left({u}^{2} + 2 u + 4\right)}$

As $\left(u - 2\right)$ cancels out and sub 2 in for u provides the final answer of $\frac{1}{12}$

Feb 8, 2017

$\frac{1}{12}$

#### Explanation:

We observe that in the present form the limit becomes $\frac{0}{0}$. Which is indeterminate.

Therefore, let us substitute
${\left(x + 8\right)}^{\frac{1}{3}} = u$
$\implies x + 8 = {u}^{3}$
$\implies x = {u}^{3} - 8$

Also as $x \to 0$, $u \to 2$

With this substitution the given question becomes

${\lim}_{u \to 2} \frac{u - 2}{{u}^{3} - 8}$

$\implies {\lim}_{u \to 2} \frac{u - 2}{\left(u - 2\right) \left({u}^{2} + 2 u + 4\right)}$
$\implies {\lim}_{u \to 2} \frac{1}{\left({u}^{2} + 2 u + 4\right)}$
$\implies \frac{1}{\left({2}^{2} + 2 \times 2 + 4\right)}$
$\implies \frac{1}{12}$