SOLUTION: when will
lim [(ax^2 + bx + c)/ (x^2 - x -6)] = 1
x->0
a) a = 0
b) a = 1
c) b = -1
d) c = -6
im getting 2 different answers so Im a bit confused...
Please can so
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-> SOLUTION: when will
lim [(ax^2 + bx + c)/ (x^2 - x -6)] = 1
x->0
a) a = 0
b) a = 1
c) b = -1
d) c = -6
im getting 2 different answers so Im a bit confused...
Please can so
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Question 551693: when will
lim [(ax^2 + bx + c)/ (x^2 - x -6)] = 1
x->0
a) a = 0
b) a = 1
c) b = -1
d) c = -6
im getting 2 different answers so Im a bit confused...
Please can some one explain it to me with solution thanks... Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! We can substitute directly because the function is continuous everywhere near 0 (the only "problems" are near -2 and 3), and obtain c = -6. We cannot claim any of the other choices, because they may or may not be true. Additionally, we cannot use L'Hopital's rule, because we need both limits to equal 0 or infinity.
