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

$\displaystyle\lim_{x\to2}\frac{x^2+4x-12}{x-2}$

Substituting $x=2$ directly yields $\dfrac{4+8-12}{2-2}=\dfrac00$. 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

$\displaystyle\lim_{x\to2}\frac{x^2+4x-12}{x-2}=\lim_{x\to2}\frac{(x-2)(x+6)}{x-2}=\lim_{x\to2}(x+6)=2+6=8$