2个回答
展开全部
1.lim dx->1 [f(x+1)-f(x)]/dx=f ' (x) ===>f(x)=f(x+1)-f ' (x)=(x^2+3x+5)-(2x+3)=x^2+x+2
同理: f(x-1)=x^2-x+1
2. cosx-cos^3 x =cosx(1-cos^2 x)=cosx sin^2 x
lim x->0 cosx-cos^3 x /X^2 =lim x->0 cosx * lim x->0 sin^2 x /x^2 =1
3 lim x->∞ (1+x2/x)^x = lim x->∞ (1+2/x)^(x/2)2=1^2=1
4 看不清
5.lim x->0- f(x)=lim x->0- sinx / x =1
lim x->0+f(x) =lim x->0+ sinx / x =1
f(0)=3
所以函数f(x)在x=0处是不连续的.
同理: f(x-1)=x^2-x+1
2. cosx-cos^3 x =cosx(1-cos^2 x)=cosx sin^2 x
lim x->0 cosx-cos^3 x /X^2 =lim x->0 cosx * lim x->0 sin^2 x /x^2 =1
3 lim x->∞ (1+x2/x)^x = lim x->∞ (1+2/x)^(x/2)2=1^2=1
4 看不清
5.lim x->0- f(x)=lim x->0- sinx / x =1
lim x->0+f(x) =lim x->0+ sinx / x =1
f(0)=3
所以函数f(x)在x=0处是不连续的.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询