Question

How do you find `lim_(x to 2) (x^3-6x-2)/(x^3+4)`, using l'Hospital's Rule or otherwise?

Answer 1

Because the expression evaluated at `x =2` does not yield an indeterminate form, (e.g. `0/0` or `oo/oo`), one cannot use L'Hospital's rule. The limit is merely the expression evaluated at `x = 2`.

Explanation:

Evaluate the expression at `x = 2`:

`lim_(x to 2) (x^3-6x-2)/(x^3+4) = (2^3-6(2)-2)/(2^3+4)`

The limit is:

`lim_(x to 2) (x^3-6x-2)/(x^3+4) = -1/2`