Question

高数 求详解？

（7）和4

Answer 1

(7)
lim(x->0) [ 3sinx+ x^2.cos(1/x) ]/ [ (1+cosx) ln(1+x) ]
=(1/2)lim(x->0) [ 3sinx+ x^2.cos(1/x) ]/ ln(1+x)
=(1/2)lim(x->0) [ 3sinx+ x^2.cos(1/x) ]/ x
=(1/2)lim(x->0) [ 3sinx/x+ x.cos(1/x) ]
=(1/2)(3+0)
=3/2
(4)
f(x)
=sinax/√(1-cosx) ; x<0
=b ; x=0
=(1/x)[lnx-ln(x^2+x) ] ; x>0
f(0-) =lim(x->0-) sinax/√(1-cosx) =lim(x->0-) ax/[(1/√2)x ] =√2.a
f(0)=b
f(0+)
=lim(x->0+) (1/x)[lnx-ln(x^2+x) ]
=lim(x->0+) (-1/x)ln [(x^2+x)/x ]
=lim(x->0+) (-1/x)ln (1+x)
=lim(x->0+) (-1/x)(x)
=-1
f(0-)=f(0) = f(0+)
√2.a =b=-1
a= -√2/2 and b=-1
(a,b)= (-√2/2 ,-1)