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What is the limit of #(x^3 -1)/(x^2+2x-3)# as x approaches 1?

Answer 1

#lim_(x->1)((x^3-1)/(x^2 + 2x - 3)) = 3/4#

In order to avoid the #0/0# indeterminate form, which you would get if you tried to evaluate this limit for #x->1#, you can a little algebraic manipulation to rewrite your initial function.

The numerator of the fraction can be factored using the formula for the difference of two cubes

#color(blue)(a^3 - b^3 = (a-b)^3 + 3ab(a-b)#

This means that you can write

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

Now focus on the denominator, which can be factored to get

#x^2 + 2x - 3 = x^2 -x + 3x - 3 = (x-1)(x+3)#

Your limit will now be

#lim_(x->1)(color(red)(cancel(color(black)((x-1)))) * [(x-1)^2 + 3x])/(color(red)(cancel(color(black)( (x-1)))) (x+3)) = lim_(x->1)((x-1)^2 + 3x)/(x+3)#

As #x->1#, this limit will be equal to

#lim_(x->1) ( (1-1)^2 + 3 * 1)/(1+3) = color(green)(3/4)#

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Answer 2

Scarlett Allison

The limit of (x^3 -1)/(x^2+2x-3) as x approaches 1 is 0.

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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.