Question #58ca2

1 Answer
Dec 20, 2017

lim_(x->oo)sinx/x=0

Explanation:

-1<=sinx<=1 AA x in RR

=>-1/x <= sinx/x<=1/x " " for all x > 0

lim_(x->oo)1/x=0

lim_(x->oo)-1/x=0

0<=lim_(x->oo)sinx/x<=0

lim_(x->oo)sinx/x=0