# Question #1488e

Oct 8, 2017

The limit does not exist.

#### Explanation:

We can't simply plug in $x = 2$, because this will give us a denominator of $0$.

However, note that at $x = 2$, the term ${x}^{\frac{1}{3}} - 2$ will NOT be $0$.
Instead, it will be ${2}^{\frac{1}{3}} - 2$, which is less than $0$.

Let's approach $x = 2$ from both sides:

Let $x = 2 + a$, where $a$ is an extremely small number. Then the negative limit is:

${\lim}_{a \to {0}^{-}} \frac{{\left(2 + a\right)}^{\frac{1}{3}} - 2}{a} = \text{negative value"/"negative value approaching 0" = "positive value"/"positive value approaching 0} = \infty$

Similarly, the positive limit is:

${\lim}_{a \to {0}^{+}} \frac{{\left(2 + a\right)}^{\frac{1}{3}} - 2}{a} = \text{negative value"/"positive value approaching 0} = - \infty$

Since the negative limit is $\infty$ and the positive limit is $- \infty$, the two limits are not the same and therefore we say the limit does not exist.

To show this limit, here is the graph of the function $y = \frac{{x}^{\frac{1}{3}} - 2}{x - 2}$
graph{(x^(1/3)-2)/(x-2) [-3.184, 6.816, -2.32, 2.68]}