Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

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We have to find the value of lim x--> -1 [(2x^2-x-3)/(x+1)]

If we substitute x = -1 in (2x^2-x-3)/(x+1) we get 0 / 0. So L'Hopital's theorem can be applied. We now have:

lim x--> -1 [(4x -1)/(1)]

substituing x = -1, we get (-4 - 1)/1

=> -5

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We have to find the value of lim x--> -1 [(2x^2-x-3)/(x+1)]

If we substitute x = -1 in (2x^2-x-3)/(x+1) we get 0 / 0. So L'Hopital's theorem can be applied. We now have:

lim x--> -1 [(4x -1)/(1)]

substituing x = -1, we get (-4 - 1)/1

=> -5

Limit = -5

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