# How do you find the limit of  (x^3 - x) / (x -1) as x approaches 1?

Aug 28, 2016

The limit can be evaluated by cancelling out $x - 1$ as follows. If you notice that ${x}^{3} - x$ has common factors of $x$, factor out $x$.

$\textcolor{b l u e}{{\lim}_{x \to 1} \frac{{x}^{3} - x}{x - 1}}$

$= {\lim}_{x \to 1} \frac{x \left({x}^{2} - 1\right)}{x - 1}$

Since ${x}^{2} - 1$ is a difference of two squares (${x}^{2} - {a}^{2}$, where $a$ is a constant), you can factor this into $\left(x + 1\right) \left(x - 1\right)$.

$\implies {\lim}_{x \to 1} \frac{x \left(x + 1\right) \cancel{\left(x - 1\right)}}{\cancel{\left(x - 1\right)}}$

$= {\lim}_{x \to 1} x \left(x + 1\right)$

Now you can just plug $x = 1$ in.

$\implies \left(1\right) \left(1 + 1\right) = \textcolor{b l u e}{2}$

And you can see from Wolfram Alpha that it is indeed correct.