What is the error in limx-2 [e^(3x^2 -12x+12)]/(x^4 -16) = 0 ?
It uses L'Hopital's rule to derive first but after doing it twice, I got 1/8

Question

What is the error in limx->2 [e^(3x^2 -12x+12)]/(x^4 -16) = 0 ?
It uses L'Hopital's rule to derive first but after doing it twice, I got 1/8

anonymous · Answer

But then the graph of it says the limit does not exist..

sid1729 · Answer

Not sure if you can apply L'Hopital's rule. As x -> 2, the numerator tends to e^0, which is 1. Since you can only apply the rule when both the numerator and denominator approach zero or +/- infinity, I don't think it applies here.