Question

Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→2 x2 + 4x − 12 x − 2

Answer 1

`\displaystyle\lim_(x\to2)(x^2+4x-12)/(x-2)`

Substituting
`x=2` directly yields
`(4+8-12)/(2-2)=\frac00`. Applying L'Hopital's rule, we have


`\displaystyle\lim_(x\to2)\frac{2x+4}1=4+4=8`

But note that
`x-2` is a factor of the numerator, which follows from the remainder theorem: if
`x=2`, then
`2^2+4(2)-12=0` which means
`x-2` is a factor of the numerator.

We have


`x^2+4x-12=(x-2)(x+6)`

and so


`\displaystyle\lim_(x\to2)(x^2+4x-12)/(x-2)=\lim_(x\to2)((x-2)(x+6))/(x-2)=\lim_(x\to2)(x+6)=2+6=8`