Question

How do you find `lim (3x^2+x+2)/(x-4)` as `x->0` using l'Hospital's Rule or otherwise?

Answer 1

`-1/2`

Explanation:

The initial form is not indeterminate. We cannot use l'Hospiital for this.

`lim_(xrarr0)(3x^2+x+2) = 2` and

`lim_(xrarr0)(x-4) = -4`, so

`lim_(xrarr0)(3x^2+x+2)/(x-4) = 2/(-4) = -1/2`