Question

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

Answer 1

`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`