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

Answer 1

#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#
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Answer 2
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#

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Answer 3

#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#.
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Answer 4

The limit of the given expression as h approaches 0 is equal to 2π.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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