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

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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 $f'(2pi)=cos(2pi)=cos(0)=1$ .

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