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>
• 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.
>
> Amir
>
>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

```

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