Question

Find BY SERIES the limit as x -> 0 of (sin x^2 - x^2)/(x^6)

Answer 1

can you write it more clearly?

Answer 2

I don't see why you would ever want to use series to solve this, you could just use l'Hopital's rule.

Answer 3

because mathematicians are picky and meticulous. It's like a "code of ethics" for mathematicians. "If there's a simple way, take the harder solution"

Answer 4

\[(\sin x ^{2} -x ^{2})/x^{6}\]

if that's your eqn. it does come out nicely...

Answer 5

Oh do they want a Taylor expansion?

Answer 6

If so that would make sense...

Answer 7

@minnie0mouse Find the Taylor expansion for sin(x^2), subtract x^2 from it and divide the whole thing by x^6. Then see what happens for x->0.

Answer 8

\[\sin(u) = u -u ^{3}/3! +u ^{5}/5! .....\]
u=x^2
\[\frac{ 1 }{ x ^{6} } * (x ^{2} - x ^{6}/3! + x ^{10}/5! ....)\]

Answer 9

@Algebraic! don't forget to subtract the x^2 before the division :)

Answer 10

whoops 1/x^4

Answer 11

(x^(-4) -1/3! + x^(4) /5! ...) -1/x^(4)

Answer 12

= -1/3! + 0 + 0 +....

Answer 13

inb4 "No, actually it was sin(x^2-x^2) and that's why I was confused!!11"

Answer 14

@Algebraic! Let's hope not :P Nice answer.

Answer 15

lol:)

Answer 16

I actually figured it out. Thank you, though! :)