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lim x->2 (8-x^3)/(x^2-4) = ?
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You have 0/0 do l'hospital's rule to get: \[\lim_{x \rightarrow 2}\frac{-3x^2}{2x}=-3\]
in any case if you plug two into the limit, you get 0/0 which is an indeterminate form, but also a form that means you can use l'hospitals rule which says you can analyze the limit by looking at the derivatives of the two functions when you have inf/ inf, 0/0 types of indeterminate limits. so taking the first derivative, we have -3x^2 / 2x, plugging 2 into the limit gives -12/4, = -3
use lhopsitals rule if you ever get 0/0 or infinity/infinity
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