limx→∞√4x2+3x−2x
I thought I could multiply both numerator and denominator by 1x, giving
limx→∞√4+3x−21x
then as x approaches infinity, 3x essentially becomes zero, so we're left with 2-2 in the numerator and 1x in the denominator, which I thought would mean that the limit is zero.
That's apparently wrong and I understand (algebraically) how to solve the problem using the conjugate, but I don't understand what's wrong about the method I tried to use.
Answer
Hint: √4x2+3x−2x=3x√4x2+3x+2x=3√4+3x+2
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