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

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Answer 1 · 2016-08-28T04:46:43

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 $x$ .

$color(blue)(lim_(x->1) (x^3 - x)/(x - 1))$

$= lim_(x->1) (x(x^2 - 1))/(x - 1)$

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

$=> lim_(x->1) (x(x + 1)cancel((x - 1)))/cancel((x - 1))$

$= lim_(x->1) x(x + 1)$

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

$=> (1)(1 + 1) = color(blue)(2)$

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

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