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Answer 1 · 2017-12-10T14:57:33

#xf(x) = o(x^(n+1))#

If #f(x) = o(x^n)# near #x=0#, this means that:

#lim_(x->0) f(x)/x^n =0#

This implies that:

#lim_(x->0) (xf(x))/x^(n+1) = lim_(x->0) f(x)/x^n =0#

so:

#xf(x) = o(x^(n+1))#

We can't establish anything in terms of #O(dot)# because:

#f(x) = o(x^n)#

tells us that #f(x)# is infinitesimal of higher order than #x^n# but does not tell us how much higher.