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

Question

Samantha Adkison · Accepted Answer

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

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#

Adam Adkins · Answer

undefined To find the limit of (x^2-3x+2)/(x^3-4x) as x approaches 2, we can substitute the value of 2 into the expression. This gives us (2^2-3(2)+2)/(2^3-4(2)). Simplifying further, we get (4-6+2)/(8-8), which equals 0/0. This is an indeterminate form. To evaluate the limit, we can use algebraic manipulation or L'Hôpital's rule. Applying L'Hôpital's rule by taking the derivative of the numerator and denominator separately, we get (2-3)/(3(2^2)-4) = -1/8. Therefore, the limit of (x^2-3x+2)/(x^3-4x) as x approaches 2 is -1/8.