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evaluate lim t approaches -3 (t^2-9)/(2t^2+7x+3)
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You need to use l'Hopitals rule for this one.
12/5
yes i know i factor out the numerator now i need help on the deonomotinor
Oh. I suppose factoring works well here.. \[2t^2 + 7t + 3 = (2t + 1)(t + 3)\] So t+3 will cancel with the t+3 on the top.
And you get (-3-3)/(2(-3)+1) = -6/-5 = 6/5
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thanks again
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