设 f(x) = ln(1+ax)/sin2x (x大于0) ,f(x) = 1 (x = 0),f(x) = (e^bx - 1)/x (x小于0)求:b

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设f(x)=ln(1+ax)/sin2x(x大于0),f(x)=1(x=0),f(x)=(e^bx-1)/x(x小于0)求:b设f(x)=ln(1+ax)/sin2x(x大于0),f(x)=1(x=0

设 f(x) = ln(1+ax)/sin2x (x大于0) ,f(x) = 1 (x = 0),f(x) = (e^bx - 1)/x (x小于0)求:b
设 f(x) = ln(1+ax)/sin2x (x大于0) ,f(x) = 1 (x = 0),f(x) = (e^bx - 1)/x (x小于0)
求:b

设 f(x) = ln(1+ax)/sin2x (x大于0) ,f(x) = 1 (x = 0),f(x) = (e^bx - 1)/x (x小于0)求:b
你的意思是已知f(x)连续要求a,b的值吧.
当x>0时,f(x) = ln(1+ax)/sin2x ,因此 x->0+ 时
limf(x)=lim[ln(1+ax)/x]/(sin2x/x),对分子分母同时求极限即知 x->0+ 时
limf(x)=a/2=1,所以a=2
当x0-时
limf(x)=lim(e^bx - 1)/x=lim(bx/x)=b,若要f(x)在0处连续则需b=1.
综上即有a=2,b=1