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+x+x∧2)+ln(1-x+x∧2)]/xsinxx→0
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/03 13:27:21
![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+x+x∧2)+ln(1-x+x∧2)]/xsinxx→0](/uploads/image/z/5435706-66-6.jpg?t=%E9%AB%98%E6%95%B0%E4%BA%8C%E6%B1%82%E6%9E%81%E9%99%90%E5%92%8C%E5%AF%BC%E6%95%B01.%E8%AE%BE4%2F%281-X%E2%88%A72%29%2Af%28x%29%3Dd%5Bf%28x%29%5D%E2%88%A72%2C%E4%B8%94f%280%29%3D0%2C%E6%B1%82f%28x%292.lim%5Bln%281%2Bx%2Bx%E2%88%A72%29%2Bln%281-x%2Bx%E2%88%A72%29%5D%2Fxsinxx%E2%86%920)
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+x+x∧2)+ln(1-x+x∧2)]/xsinxx→0
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+x+x∧2)+ln(1-x+x∧2)]/xsinx
x→0
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+x+x∧2)+ln(1-x+x∧2)]/xsinxx→0
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)=0
2.∵[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
1.
右边 d[f(x)]∧2=2 f(x) df(x)
所以
4/(1-X∧2)*f(x)=2 f(x) df(x)
2/(1-X∧2)=df(x)
f(x)=积分(从0到x) dt 2/(1-t∧2)=2 arctan x
2.lim[ln(1+x+x∧2)+ln(1-x+x∧2)]/xsinx
罗毕达法则,分子分母同时求导
分子...
全部展开
1.
右边 d[f(x)]∧2=2 f(x) df(x)
所以
4/(1-X∧2)*f(x)=2 f(x) df(x)
2/(1-X∧2)=df(x)
f(x)=积分(从0到x) dt 2/(1-t∧2)=2 arctan x
2.lim[ln(1+x+x∧2)+ln(1-x+x∧2)]/xsinx
罗毕达法则,分子分母同时求导
分子求导=(2 (x + 2 x^3))/(1 + x^2 + x^4)
在x接近0时分子约等于 2 x
分母求导=x cos x + sin x
在x接近0时分母约等于 x+x=2x (cos x~1 sin x~x)
所以极限等于 (2x)/(2x)=1
收起