Limit of a power series representation

s3a

Nov 2008
624
5
I know how to do everything except for the limit part.

(My work and the question are attached)

So if somone could please explain to me what to do for that part, I would greatly appreciate it as always!

Thanks in advance!
 

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Last edited:
Oct 2009
4,261
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I know how to do everything except for the limit part.

(My work and the question are attached)

So if somone could please explain to me what to do for that part, I would greatly appreciate it as always!

Thanks in advance!

I read carefully the note you attached (it wasn't that nice...(Worried) ) and I couldn't find anywhere a limit...

Tonio
 

s3a

Nov 2008
624
5
That's because it's the limit part that I don't know what to do. (That's what I'm asking)
 
Oct 2009
4,261
1,836
That's because it's the limit part that I don't know what to do. (That's what I'm asking)

Either you or I (or both) aren't paying attention: nowhere in that piece of paper you attached appears a limit...what do you want?!! The limit of the last line when \(\displaystyle x\to o\) ? Or do you mean that some of the infinite series that appear there is to be considered as the limit of their partial sums.... you need the limit of what !?

Tonio
 

s3a

Nov 2008
624
5
lim x->0 (the pdf says so - not the jpeg that I scanned of my paper)
 
Oct 2009
4,261
1,836
lim x->0 (the pdf says so - not the jpeg that I scanned of my paper)

Oh ,finally! Who checks the PDF if you send an attached note? Anyway, open up the right-hand series in the last line:

\(\displaystyle \frac{1}{2}\sum^\infty_{n=2}(-1)^n\frac{x^{2n-4}}{(2n)!}=\frac{1}{2}\left(\frac{1}{4!}-\frac{x^2}{6!}+\ldots\right)\) . As you can see, all the summands on the right hand but the first one contain positive powers of x and thus vanish

when \(\displaystyle x\to 0\) ,and thus the limit is \(\displaystyle \frac{1}{2}\cdot\frac{1}{24}=\frac{1}{48}\) which, by the way, agrees with L'Hospital...(Wink)

Tonio
 
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s3a

Nov 2008
624
5
Oh my god! That's so clever! Because 0^0 does not exist but the limit as x->0 to the exponent 0 is 1! Also isn't it a convention to have 0^0 = 1 in sums?
 
Oct 2009
4,261
1,836
Oh my god! That's so clever! Because 0^0 does not exist but the limit as x->0 to the exponent 0 is 1! Also isn't it a convention to have 0^0 = 1 in sums?

Not that I know...it could be agreed on, though.

Tonio
 

s3a

Nov 2008
624
5
Ok but the limit of x->0 to the exponent 0 IS 1, right? (and that's why the answer I get is correct)
 
Apr 2010
384
153
Canada
Ok but the limit of x->0 to the exponent 0 IS 1, right? (and that's why the answer I get is correct)
The limit of any real number to the exponent 0 is 1, and the limit of 0 to 0 is an indeterminent form and we need to use L'Hopitals here.

In some cases it is acceptable to let \(\displaystyle 0^0 = 1 \) but this depends on the context of the problem, and is not always the correct subsitition. Other times it is correct to keep it as an indeterminent form.

Math Forum: Ask Dr. Math FAQ: Zero to Zero Power