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Find the limit as x approaches infinity of #y=sinx/x#?

Question

Find the limit as x approaches infinity of #y=sinx/x#?
Find the limit as x approaches infinity of #y=sinx/x#?

Answers

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

Explanation:

You're going to want to use the squeeze theorem for this. Recall that #sinx# is only defined on #-1 ≤ sinx ≤ 1#. Therefore

#-1/x ≤ sinx/x ≤ 1/x#

And since #lim_(x-> oo) -1/x = lim_(x-> oo) 1/x = 0#, then #lim_(x->oo) sinx/x = 0#.

Hopefully this helps!


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