Re: Calculus : limits

*To*: mathgroup at smc.vnet.net*Subject*: [mg51312] Re: [mg51279] Calculus : limits*From*: DrBob <drbob at bigfoot.com>*Date*: Thu, 14 Oct 2004 06:35:47 -0400 (EDT)*References*: <200410120557.BAA19198@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

That's the right answer, and a Plot confirms it at my machine. f[x_] = (Sin[x] - Sin[2*x])/x; Limit[f[x], x -> 0] -1 Limit[Abs[f[x]], x -> 0] 1 Plot[f[x], {x, -1, 1}] A look at the Series representations makes the answer very clear: Series[Sin[x], {x, 0, 5}] Series[Sin[2*x], {x, 0, 5}] (%% - %)/x SeriesData[x, 0, {1, 0, -1/6, 0, 1/120}, 1, 6, 1] SeriesData[x, 0, {2, 0, -4/3, 0, 4/15}, 1, 6, 1] SeriesData[x, 0, {-1, 0, 7/6, 0, -31/120}, 0, 5, 1] Or, in even simpler terms, when x is close to 0, Sin[x] is close to x and Sin[2x] is close to 2x, so their difference is close to -x. Divide by x, and that's close to -1. Take Abs and you get 1. Bobby On Tue, 12 Oct 2004 01:57:42 -0400 (EDT), Amir <z64043 at netscape.net> wrote: > Hi, > > I'd like to find the limit of > Limit[(Abs[Sin[x]-Sin[2 x]]) / x, x->0] > > I use Mathematica v.5. I get the wrong (??) answer : 1 > > While I try to display the graph of this function by using "Plot", it > seems that there is no limit at the point x=0. > Please help... > > Amir > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Calculus : limits***From:*Amir <z64043@netscape.net>