How do you find the limit of #sinx / x# as x approaches #oo#?

1 Answer
A08
Mar 29, 2016

We use squeeze theorem, which says that for three functions #g(x), f(x), and h(x)#,
If #g(x)<= f(x) <= h(x), and lim_ (x->a)g(x)=lim_ (x->a)h(x)=L#
then #lim_ (x->a)f(x)=L#.

#lim_(x->+-oo)sin x/x=0#

Explanation:

Given that #f(x)=sinx/x#
We know that #sinx# oscillates between #-1 and +1# for all values of #x#. Therefore, if we set #g(x)=(-1)/xand h(x)=(+1)/x# we have located the two functions satisfying the first condition that
#g(x)<= f(x) <= h(x)# which can be written as

Given that #(-1)/x<= sinx/x <= (+1)/x "for all values of " x " in " (-oo,oo)#
We know that #lim_(x->+-oo)(-1)/x=0 and "also that" lim_(x->+-oo)(+1)/x=0#.

It follows from the squeeze theorem that
#lim_(x->+-oo)sin x/x=0#

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