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lim x>0+ (1 + 3x/2)^1/x
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gotta be \(e^{\text{something}}\) my guess is \(e^{\frac{3}{2}}\) since \[\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e\]
you have a choice here one way it to take the log and get \[\frac{1}{x}\ln(1+\frac{3x}{2})\] then take the limit of that using l'hopital, then exponentiate
second and actually correct (also equivalent way) is to write \[(1+\frac{3x}{2})^{\frac{1}{x}}=e^{\frac{1}{x}\ln(1+\frac{3x}{2})}\] and then take the limit, but all with work will be identical
thanx.. the second one will probably work.. thanx so much
yw
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