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.