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Mathematics 19 Online
OpenStudy (anonymous):

lim x^3+ x^2- 5x+3/ x^3-3x+2 x-->1

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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