设y=(1+3tanx²)^(1/x)
lny=ln(1+3tanx²)/x
lim0> ln(1+3tanx²)/x
=lim 6x/((cos²x²)*(1+3tanx²))
=lim 6x/(cos²x²+3sinx²*cosx²)=0/1=0
所以
lim (x->0) (1+3tanx²)^(1/x)=lim (x->0) y
=lim (x->0)e^(lny)=e^0=1指数是1/x的平方。。。设y=(1+3tanx²)^[(1/x)^2]lny=ln(1+3tanx²)/x²lim0> ln(1+3tanx²)/x²=lim 3/((cos²x²)*(1+3tanx²))=lim 3/(cos²x²+3sinx²*cosx²)=3/1=3所以lim (x->0) (1+3tanx²)^(1/x)=lim (x->0) y=lim (x->0)e^(lny)=e³