[tex]\displaystyle\lim_{x\to2}\frac{x^2+4x-12}{x-2}[/tex]
Substituting [tex]x=2[/tex] directly yields [tex]\dfrac{4+8-12}{2-2}=\dfrac00[/tex]. Applying L'Hopital's rule, we have
[tex]\displaystyle\lim_{x\to2}\frac{2x+4}1=4+4=8[/tex]
But note that [tex]x-2[/tex] is a factor of the numerator, which follows from the remainder theorem: if [tex]x=2[/tex], then [tex]2^2+4(2)-12=0[/tex] which means [tex]x-2[/tex] is a factor of the numerator.
We have
[tex]x^2+4x-12=(x-2)(x+6)[/tex]
and so
[tex]\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[/tex]
