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 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!
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
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...
Tonio
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