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Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim h → 0 ((−9 + h)2 − 81)/h
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Have you considered expanding the square and simplifying the numerator?
Yes. I get -18+x over x
plugging in zero would make it DNE ?
1) "Plug in" doesn't mean anything. Never do that. 2) Where did 'x' come from? 3) Try again. (−9 + h)2 − 81 81 - 18h + h^2 - 81 h^2 - 18h Now, put the fraction back together and see if anything magical happens.
h^2-18h over h^2
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Can I simply more?
just over \(h\) dear then divide
\[\frac{h^2-18h}{h}\] factor the \(h\) out of the numerator and cancel, or else divide each term by \(h\) same thing THEN plug in 0 for \(h\)
i get h-18
-18 is the limit?
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NOW what happens as h disappears?
yes
thanks
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