this isn't so much a *problem* as a well, problem ðŸ˜› it came up in maths the other day, and my teacher anounced that e had no idea why this was, and neither did anyone else in the maths department...and i have no idea what you mean about power series and sin ðŸ˜•

Originally posted by royalchicken Hey genius. d(sin 2h)/dh = 2cos2h. Since sin2h--->0 as h--->0, and h--->0 as h--->0, just apply l'Hopital's rule and you get:

Indeed you can. Good show. I think this was mentioned before, and indeed I prefer to think of the circular functions in terms of their Maclaurin series, because for complex arguments the standard definition of say, sin z becomes rather meaningless.

You know guys, sometimes I think you just post bogus stuff to make everyone else that comes in here feel so stupid! ðŸ™„ Hahahaha, all these rules I have never heard of and all these nuts equations! My brain starts wimpering on the spot when I see that stuff! Hahahaha, Just Kidding guys. I'm actually jealous that I can't help out with this stuff! Carry on. ðŸ˜€