       Re: Unevaluated limit

• To: mathgroup at smc.vnet.net
• Subject: [mg84513] Re: [mg84505] Unevaluated limit
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Mon, 31 Dec 2007 21:17:29 -0500 (EST)
• References: <8036743.1199091028231.JavaMail.root@m08>

```Here are a couple of approximations via Series:

cos[x_] = Normal@Series[Cos[x], {x, 0, 4}]

1 - x^2/2 + x^4/24

g[x_] = Normal@
Series[2*x^2*Exp[6 x]*(1 - Cos[Exp[-3 x]*Tan[2/x]]), {x, Infinity,
4}]

\[ExponentialE]^(
6 x) (2 x^2 - 2 x^2 Cos[\[ExponentialE]^(-3 x) (8/(3 x^3) + 2/x)])

f[x_] = Expand[g[x] /. Cos -> cos]

4 - (1024 \[ExponentialE]^(-6 x))/(243 x^10) - (
1024 \[ExponentialE]^(-6 x))/(81 x^8) - (
128 \[ExponentialE]^(-6 x))/(9 x^6) + 64/(9 x^4) - (
64 \[ExponentialE]^(-6 x))/(9 x^4) + 32/(3 x^2) - (
4 \[ExponentialE]^(-6 x))/(3 x^2)

The second Series (for g) applies Cos to (VERY) small arguments when x is
large, so the first Series (for cos) is a good approximation there.

The result is f, which clearly approaches 4 as x->Infinity.

Bobby

On Mon, 31 Dec 2007 01:57:59 -0600, Liverpool <x at y.z> wrote:

> I have to evaluate this limit (I know the answer is 4)
>      Limit[2*x^2*Exp[6 x]*(1 - Cos[Exp[-3 x]*Tan[2/x]]), x -> Infinity]
>
> but the output is unevaluated:
>
>      Limit[2 \[ExponentialE]^(6 x) x^2 (1 - Cos[\[ExponentialE]^(-3 x)
> Tan[2/x]]), x -> \[Infinity]]
>
> Is there any way to evaluate it symbolically?
>
> Thanks, Liverpool
>
>
>
>

--

DrMajorBob at bigfoot.com

```

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