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How do you find the limit of #(x^3 - x)/(x-1)^2# as x approaches 1?

Answer 1

#lim_(x->1-)[(x^3-x)/(x-1)^2]=-oo#
#lim_(x->1+)[(x^3-x)/(x-1)^2]=+oo#

#lim_(x->1)(x^3-x)/(x-1)^2=(x(x+1)(x-1))/(x-1)^2=(x^2+x)/(x-1)=2/0=oo#

However when #lim_(x->1-)[(x^3-x)/(x-1)^2]=-oo# and #lim_(x->1+)[(x^3-x)/(x-1)^2]=+oo#. This is because you can approach #1# at either side and this is important because it will determine the sign of #x-1#.

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Answer 2

Charles Anctil

To find the limit of (x^3 - x)/(x-1)^2 as x approaches 1, we can use algebraic manipulation and the concept of limits. By factoring the numerator, we get (x)(x^2 - 1). Simplifying the denominator gives (x-1)(x-1). Canceling out the common factor of (x-1), we are left with (x^2 + 1). As x approaches 1, the expression (x^2 + 1) approaches 2. Therefore, the limit of (x^3 - x)/(x-1)^2 as x approaches 1 is 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.