Question

Using Maclaurin series to find a limit. See attachment.

Answer 1

I'm getting 0/0. But, that doesn't seem to be correct. Any ideas?

Answer 2

lets define the series first ...

Answer 3

\[e^u=\sum_0~\frac1{n!}u^n\]
\[e^{(-3x^3)}=\sum_0~\frac1{n!}(-3x^3)^n\]
dividing htat by 8x^9 gets us
\[\frac{e^{(-3x^3)}}{8x^9}=\sum_0~\frac1{8(n!)}\frac{(-3x^{2})^n}{x^9}\]
then its just a matter of popping out the parts up relating to the x^6 term to play with the other parts

Answer 4

or pop them out before the division might clean it up a little

Answer 5

\[\frac1{0!}(-3x^3)^0=1\]\[\frac1{1!}(-3x^3)^1=-3x^3\]\[\frac1{2!}(-3x^3)^2=\frac{9}{2}x^6\]

Answer 6

What I did was substitute e^(-3x^3) with the attached image, and then everything cancelled out in the numerator.

Answer 7

right, so we are left with:\[\frac{1}{8}\sum_{3}\frac{1}{n!}(-3)^n~x^{3n-9}\]
\[\frac{3}{8}\sum_{3}\frac{(-1)^n}{n!}~x^{3(n-3)}\]

Answer 8

this has a non-x term to start with so its the rest that zeros out

Answer 9

\[\frac{3}{8}\sum_{3}\frac{(-1)^n}{n!}~x^{3(n-3)}\]
\[\frac{3}{8}(\frac{-1}{3!}+\sum_{4}\frac{(-1)^n}{n!}~x^{3(n-3)})\]
\[\frac{3}{8}(\frac{-1}{6})=\frac{-1}{16}\]

Answer 10

im sure i made a typo in there somewhere

Answer 11

Yeah, that doesn't seem to be the answer either.

Answer 12

-9/16 is the answer, which means i lost a 3 along the way

Answer 13

ahh, i pulled out the 3 instead of using it

Answer 14

(-3)^n is NOT the same as (-1)^n 3

Answer 15

\[\frac{1}{8}\sum_{3}\frac{1}{n!}(-3)^n~x^{3n-9}\]
\[\frac{1}{8}\sum_{3}\frac{(-1)^n~3^n}{n!}~x^{3n-9}\]
\[\frac{1}{8}\left(\frac{-3^3}{3!}+\bcancel{\cancel{\sum_{4}\frac{(-1)^n~3^n}{n!}~x^{3n-9}}}\right)\]

Answer 16

Yes, that's correct. Now I just need to look over your processes, because I did it a little differently.

Answer 17

Thanks for your help! I still need to process how you did it all, but I don't have time right now.

Answer 18

i had to process it as a worked it :) youll do fine