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

Jun 8, 2016

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

#### Explanation:

$\sin \left(2 \pi + h\right) = \sin \left(h\right)$ periodic function with period $2 \pi$
$\sin \left(2 \pi\right) = 0$

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

Hence the above limit is in the form of $\frac{0}{0}$ we can apply L'Hopital rule hence

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

Jun 9, 2016

$1$

#### Explanation:

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

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

So, if $f \left(x\right) = \sin \left(x\right)$, we see that the derivative of sine is

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin \left(x + h\right) - \sin \left(x\right)}{h}$

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

$f ' \left(2 \pi\right) = {\lim}_{h \rightarrow 0} \frac{\sin \left(2 \pi + h\right) - \sin \left(2 \pi\right)}{h}$

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