高数二求极限和导数1.设4/(1-X∧2)*f(x)=d[f(x)]∧2,且f(0)=0,求f(x)2.lim[ln(1

1。∵4f(x)/(1-x²)=d[f²(x)]==>4f(x)/(1-x²)=2f(x)d[f(x)]==>f(x){d[f(x)]-2/(1-x²)}=0∴d[f(x)]-2/(1-x²)=0,或f(x)=0（1）当d[f(x)]-2/(1-x²)=0时,有d[f(x)]=2/(1-x²)==>f(x)=∫2dx/(1-x²)=∫[1/(1+x)+1/(1-x)]dx=ln│1+x│+ln│1-x│+C (C是积分常数)=ln│(1+x)/(1-x)│+C∵f(0)=0 ==>C=0∴f(x)=ln│(1+x)/(1-x)│（2）显然f(x)=0是满足条件f(0)=0的解综合（1）（2）知,f(x)=ln│(1+x)/(1-x)│,或f(x)=02。∵[ln(1+x+x²)+ln(1-x+x²)]/(xsinx)=ln[(1+x+x²)(1-x+x²)]/(xsinx)=ln(1+x²+x^4)/(xsinx)=(x/sinx)*[ln(1+x²+x^4)/x²]又lim(x->0)(x/sinx)=1/[lim(x->0)(sinx/x)]=1 (∵lim(x->0)(sinx/x)=1)lim(x->0)[ln(1+x²+x^4)/x²]=lim(x->0)ln[(1+x²+x^4)^(1/x²)]=ln{lim(x->0)[((1+x²+x^4)^(1/(x²+x^4)))(1+x²)]=ln{lim(x->0)[e^(1+x²)]} (应用重要极限lim(x->0)[(1+x)^(1/x)]=e)=lne=1∴原式=lim(x->0)(x/sinx)*lim(x->0)[ln(1+x²+x^4)/x²]=1*1=1