Not sure I understand your mail. Your original mail said that the correct
answer is 0 (which I agree with). Now you're saying that the correct
answer is not zero? I'm confused.
The theorem you cite isn't applicable in this case, because we don't know
that limit(f(x),x,0) exists. We do however know that f(x) is bounded,
which is good enough, because limit(f(x)*h(x),x,0) = 0 if f(x) is bounded
in a neighborhood of 0 and limit(h(x),x,0)=0.
For example, limit(sin(1/x)*x,x,0) = 0 because sin(1/x) is bounded, though
it doesn't have a limit at 0.
-s
On Thu, Mar 28, 2013 at 12:46 PM, Thomas D. Dean <tomdean at speakeasy.org>wrote:
> On 03/28/13 06:57, Stavros Macrakis wrote:
>
>> I don't believe there's any way of telling the limit subsystem that an
>> arbitrary function of the limit variable is bounded. Sorry.
>>
>
> I think limit returns the correct value, the given function.
>
> Maple returns 0, which, I believe is not correct.
>
> Theorem: limit(f(x)*g(x),x,a) = limit(f(x),x,a) * limit(g(x),x,a), if
> both limits exist;
>
> but, if limit(f(x),x,a) = infinity, then the limit of the product is
> undefined.
>
> Correct?
>
>
> Tom Dean
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