• anonymous
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
  • katieb
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  • anonymous
But then the graph of it says the limit does not exist..
  • sid1729
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.

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