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...
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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