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Evaluate the limit. Lim x-->0- (x^2-8)/x??
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\[\lim_{x\to 0^{-}}\frac{x^2-8}{x}=\infty\]
how do you know?
like can you explain?
bottom is getting very small (but is negative)...top is getting close to -8. (so fraction is going to +infty since... neg/neg=pos
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