How do you find the limit of #(x^2-3x+2)/(x^3-4x)# as x approaches 2?

1 Answer
Andrea S.
Nov 30, 2016

#lim_(x->2) frac (x^2-3x+2) (x^3-4x) = 1/8#

Explanation:

Factorize the numerator and denominator:

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

#x^3-4x = x(x-2)(x+2)#

You can see you can simplify the rational function, making it continuous in #x=2#:

#frac (x^2-3x+2) (x^3-4x) = frac ((x-1) cancel((x-2))) (x cancel((x-2))(x+2)) = frac (x-1) (x(x+2))#

As in this form the function is continuous, the limit equals the value:

#lim_(x->2) frac (x-1) (x(x+2)) = frac (2-1) (2(2+2)) = 1/8#

Alternatively, for instance in the case of polynomials that are harder to factorize, you can use L'Hôpital's rule, stating that as the limit is in the indeterminate form #0/0# and as numerator and denominator are differentiable around #x=2#,

#lim_(x->2) frac (x^2-3x+2) (x^3-4x) = lim_(x->2) frac (d/dx(x^2-3x+2)) (d/dx(x^3-4x))#

#lim_(x->2) frac (x^2-3x+2) (x^3-4x) = lim_(x->2) frac (2x-3) (3x^2-4) = frac (2 * 2 -3) (3 *2^2-4) =1/8#

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