Originally Posted by

**javax** If you didn't solve yet, just like these guys told, I did like this and I think it's ok:

$\displaystyle sin(x+y) = sinxcosy + cosxsiny $

$\displaystyle \lim_{x\to 0}{\frac{sinx}{x}} = 1$

then

$\displaystyle \lim_{h\to 0}{\frac{sin(\pi+h)-sin\pi}{h}} = \lim_{h\to 0}{\frac{sin\pi cosh+cos\pi sinh - sin\pi}{h}}$

$\displaystyle = {\color{red} \lim_{h\to o}{\frac{sin\pi(cosh-1)+cos\pi sinh}{h}}} = \lim_{h\to 0}{\frac{cos\pi sinh}{h}} = cos\pi \lim_{h\to 0}{\frac{sinh}{h}} = cos\pi = -1$