Mark M. answered 09/18/16
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Retired math prof. Very extensive Precalculus tutoring experience.
I don't understand what you mean by the first one, but here are solutions for the other two:
limx→∞[ex/(1-ex)] has the indeterminate form ∞/(-∞)
= limx→∞[1/((1/ex) - 1)] = 1/(0-1) = -1
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limx→-∞[x2/√(2x4-3x)] has the indeterminate form ∞/∞
√(2x4-3x) can be written as √(x4)√(2-(3x/x4)) = x2 √(2 - (3/x3))
So, the given limit is equivalent to limx→∞[1/√(2 - (3/x3))] = 1/(√2-0)
= 1/√2