#lim_(h->0) [sin(2π+h) - sin(2π)]/h =# ?

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Answer 1 · 2016-06-08T16:58:14

#lim_(h->0) [sin(2π+h) - sin(2π)]/h = 1#

#sin(2pi+h)=sin(h)# periodic function with period #2pi#
#sin(2pi)=0#

#lim_(h->0) [sin(2π+h) - sin(2π)]/h = lim_{h->0}sin(h)/h = 1#

Answer 2 · 2016-06-08T17:28:42

Hence the above limit is in the form of #0/0# we can apply L'Hopital rule hence

#lim_(h->0) ((sin(2pi+h)-sin(2pi))')/[(h)']= lim_(h->0) cos(2pi+h)=cos2pi=1#

Answer 3 · 2016-06-09T15:15:21

#1#

This question is very similar to the the limit definition of the derivative:

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

So, if #f(x)=sin(x)#, we see that the derivative of sine is

#f'(x)=lim_(hrarr0)(sin(x+h)-sin(x))/h#

Since instead of #x# we have #2pi#, this is evaluating the derivative of sine at #2pi#.

#f'(2pi)=lim_(hrarr0)(sin(2pi+h)-sin(2pi))/h#

Since #f(x)=sin(x)#, we know sine's derivative is #f'(x)=cos(x)#, so #f'(2pi)=cos(2pi)=cos(0)=1#.