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Find the limit of x^2-4/x^3+8 as x approaches -2 ?
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I am assuming you have \[\frac{ ( x ^{2}-4) }{ (x^{3}+8) }\] in which case you will need to factor the top and the bottom (difference of squares for the top and sum of cubes for the bottom) which will get you \[lim_{ X \rightarrow -2} \frac{ ( x-2)(x+2) }{ (x+2)(x^2-2x+4) }\] nd then you can factor the x+2 out and plug in -2 for the answer
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