How do you determine the limit of #sin(x)/x# as x approaches infinity?

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Answer 1 · 2018-04-08T17:48:59

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

We'll use the Squeeze Theorem, which states that if we have some function #f(x),# we define new functions #h(x), g(x)# such that

#h(x)<=f(x)<=g(x)#

Then, we'll take #lim_(x->a)h(x), lim_(x->a)g(x).#

If these limits are equal, then #lim_(x->a)f(x)=lim_(x->a)h(x)=lim_(x->a)g(x)#

Or, all the limits are equal.

So, recall that

#-1<=sinx<=1#

Then,

#-1/x<=sinx/x<=1/x#

Take #lim_(x->oo)-1/x, lim_(x->oo)1/x:#

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

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

So,

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