Question

Taylor series limit, could you guys help out? Check the attachment.

Answer 1

It has to be done with Taylor series, I'm going to show what I've done so far... (p.s: you obviously don't know it's equal to -1/6)

Answer 2

Okay... denominator there's nothing much to transform, so I'm going to work on the numerator, beginning from the fact that I didn't really get "when to stop". My teacher said, you have to consider the smallest polinomial, since when x approaches 0 that's the one who's "bossing around" (lol)... okay.. let's start...

\[\log(1+t) = t- \frac{t^2}{2} + \frac{t^3}{3} -\frac{t^4}{4} + o(t^5)\]
Okay dokie, now t = -x^2...
\[ -x^2 - \frac{(-x^2)^2}{2} + \frac{(-x^2)^3}{3} -\frac{(-x^2)^4}{4} + o(x^{10})\]
\[ -x^2 - \frac{-x^4}{2} + \frac{-x^6}{3} +\frac{x^8}{4} + o(x^{10})\]

Now... x sinx!

\[x(x-\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +o(x^8))\]
Which goes under the name of...
\[x^2-\frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} +o(x^8)\]

Now I put everything togetha', getting.....

\[\frac{x^2-\frac{x^4}{3!} + \frac{x^6}{5!} - {x^8}{7!} + o(x^8) -x^2+\frac{x^4}{2}-\frac{x^6}{3}+\frac{x^8}{4} +o(x^{10})}{x^2(2x+x^2)^2}\]

You guys follow 'till here?

Answer 3

Correction!
\[\large \frac{{{x^2} - \frac{{{x^4}}}{{3!}} + \frac{{{x^6}}}{{5!}} - \frac{{{x^8}}}{{7!}} - {x^2} - \frac{{{x^4}}}{2} - \frac{{{x^6}}}{3} - \frac{{{x^8}}}{4} + o({x^8})}}{{{x^2}{{(2x + {x^2})}^2}}}\]

Answer 4

lol look at ur face

Answer 5

I know right? XD But I need help with this thingy... are you able to lend a hand ? O_O

Answer 6

not me, sorry, where'd myininaya go?

Answer 7

I have no idea... any way to call him/her? :P. I just started out yesterday with Taylor's series and I didn't completely get it.

Answer 8

She can be found on the dashboard. Hold the cursor over their name and it tells you what problem they are viewing (if any). Her, JamesJ, Zarkon, and a few others are your best bet, but others come along who know this stuff. I know taylor series but not limits like this...

Answer 9

i see a mistake

Answer 10

where ?

Answer 11

\[-x^2 - \frac{x^4}{2} + \frac{-x^6}{3} -\frac{x^8}{4} + o(x^{10})\]

Answer 12

for the log(1-x^2) part

Answer 13

Yes, I corrected it in my third post, didn't I? :O

Answer 14

i don't know i haven't got that far lol

Answer 15

I didn't get when I am supposed to stop writing polynomials.
As long as I understood (from my university teacher) I have to find the easiest polynomial and check the degree. Then write as many polynomials I need in order to "reach" that degree. I might be mistaken because, as I said, I didn't really get it at all >.<

Answer 16

Okay, I am following, what should I do now, sum and collect like terms in order to simplify with the denominator? I tried doing it before and it resulted in a O_O (@my pic)

Answer 17

i think i need to write it all out on paper give me a few

Answer 18

Ok thanks a lot for your time that's appreciated ! ^_^

Answer 19

here is a clue
\[(2x+x^2)^2=(x[2+x])^2=x^2(2+x)^2\]

Answer 20

this is the key

Answer 21

so we get \[\frac{-(\frac{1}{2}-\frac{1}{6})+blah}{(2+x)^2}\]

Answer 22

\[x->0=> blah->0\]
so we have

\[-\frac{\frac{1}{2}-\frac{1}{6}}{(2)^2}\]

Answer 23

\[-\frac{3-1}{6(2^2)}\]

Answer 24

\[\frac{-2}{6(2^2)}=\frac{-1}{6(2)}=\frac{-1}{12}\]
so we get something a little differ there must be a mistake somewhere

Answer 25

by the way i grouped my like terms together
i have a thanksgiving thing to go to
sorry

Answer 26

Aw. Okie dokie. :) Thanks anyway, you've been of great help. I'll post the solution tomorrow as soon as I get my head into the book! :)

Answer 27

okay, everything was fine, just at the denominator it was x^4(2+x)^2 so everything simplifies and it goes...
-2/3 * 1/4 = -1/6.

About "when to stop", the smartest thing I should have done was to factor out the denominator first, if I pulled out the x^4 in the very beginning, I would have understood that all them x^6 and x^8 are pretty un-necessary, since they go to zero anyway, I just needed a x^4 so I could have canceled out.
Thanks for your help! I appreciate it ! :)

Answer 28

oh i see my mistake it should have been
\[-\frac{\frac{1}{2}+\frac{1}{6}}{(2)^2}\]