Question

lim x→0 [log(1-x^9)]/[sinx^3 - x^3]. hint taylor.maclaurin. should i do like f=f(0)+f'(0)x+(1/2)f''(0)x^2......it really hard. other ways

Answer 1

sorry its hard to see. i dont know how to make it look better. please teach me how.

Answer 2

Is this the limit?:\[lim_{x→0} \frac{log(1-x^9)}{sin(x^3) - x^3}\]

Answer 3

Yes and the question hint is taylor or maclaurin

Answer 4

For the LaTeX, I used: \ [lim_{x→0} \frac{log(1-x^9)}{sin(x^3) - x^3}\]
(without the space after the first backslash)

Answer 5

I'll try that when on pc.

Answer 6

The key will be that after only a few derivatives of the denominator you get a term which isn't zero when x -> 0.

Answer 7

I need to leave now. I'll get back later.

Answer 8

Okay, I was wrong about it being a few terms. I can't avoid L'Hospital here. Either you need to take the ninth derivative of both the top and the bottom, or get pretty clever otherwise.
\[g(x) = \sin(x^3)-x^3\]
\[g'(x) = 3x^2\cos(x^3)-3x^2\]
\[g''(x) = -9x^4\sin(x^3)+6x\cos(x^3)-6x\]...
\[g^9(x) = ...-60,480\cos(x^3)\]
Similarly,
\[f(x) = \log{(1-x^9)}\]
\[f^9(x) = -9! = -362880\]
\[\frac{f^9(0)}{g^9(0)}=\frac{-362880}{-60480}=6\]

Answer 9

this sure take a lot of time in exam. this way is impossible for me. but it really true

Answer 10

Yeah, I don't know how to think about shortcutting the denominator. I can see now how it is (somewhat) easy to see that the ninth derivative of the numerator resolves to -9! when x is zero.

Answer 11

anyway thank you, you give me idea.