How do you find #lim sintheta# as #theta->oo#?

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1 Answer
NickTheTurtle
Jul 25, 2018

Does not exist

Explanation:

#sin(theta)# oscillates in between #-1# and #1# as #theta# approaches #oo#. In other words, it does not converge to a single value.

One way to show this is through the definition of a limit to positive infinity, which states that, if #lim_(x->oo)f(x)=L#, then for any given #epsilon>0#, however small, there exists some #c# such that #L-epsilon<= f(x)<= L+epsilon# for all #x>c#.

This is clearly impossible for #lim_(theta->oo)sin(theta)#, as no matter how large #theta# is, #sin(theta)# will still oscillate between #-1# and #1#. Thus, the limit does not exist.

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