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OpenStudy (anonymous):
lim x^3+ x^2- 5x+3/ x^3-3x+2 x-->1
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OpenStudy (anonymous):
are you allowed l'hospital ?
OpenStudy (anonymous):
how would I solve it though?
OpenStudy (anonymous):
lim [3x^2+2x-5]/[3x^2-3] , by l'HOSPITAL rule lim (6x+2)/(6x) , by l'Hospital rule again now let x-->1, equals 4/3. Result is 4/3
OpenStudy (anonymous):
thank you very much!
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