f(x)在正负无穷内可倒,且在x→∞时 limf '(x)=e,lim[ (x+c)/(x-c)]^x=lim[f(x)-f(x-1)],求c.

问题描述:

f(x)在正负无穷内可倒,且在x→∞时 limf '(x)=e,lim[ (x+c)/(x-c)]^x=lim[f(x)-f(x-1)],求c.
我书上答案题的提示是:拉格朗日中值定理和第二重要极限公式!最后答案等于 c=1/2

根据拉格朗日中值定理,lim(x→∞)(f(x)-f(x-1))=lim(x→∞)f'(w),x-1lim((x+c)/(x-c))^x=lim((1+c/x)/(1-c/x))^x=(lim(1+c/x)^x) ÷(lim(1-c/x)^x)=e^c/e^(-c)=e^(2c)
故c=1/2