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What is #lim_(xtooo)(sin2x)/x#?

1 Answer

Apr 13, 2018

#0#

Explanation:

Think of this as:
#lim_(xtooo)sin2x*lim_(xtooo)1/x#

As #x# tends to #oo#, #1/x# gets smaller and smaller, until it is effectively #0#

However, #sin(2x)# can be be anywhere in the range #[-1,1]# at infinity.

So, we have: #lim_(xtooo)sin2x*lim_(xtooo)1/x=[0(-1),0(1)]=0#

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