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Re: Unevaluated limit

"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?

Yes, although it's not "direct":

In[4]:= Simplify[Normal[Series[
2*x^2*Exp[6 x]*(1 - Cos[Exp[-3 x]*Tan[2/x]]), {x, Infinity, 0}]]]

Out[4]= 4*E^(6*x)*x^2*Sin[1/(E^(3*x)*x)]^2

In[5]:= Limit[%, x -> Infinity]

Out[5]= 4

Perhaps someone else will have a better way.


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