Question

Done

Answer 1

The part I can't do:

Show that there exists a d>0 such that for all x with
0<mod(x)<d, g(x) i not 0, and prove that the limit of f(x)/g(x) is a(sub)n/b(sub)n as x goes to zero.

Answer 2

What's
\[C^{(n+1)}([a,b])\]?

Answer 3

sorry, that means that it is n+1 times differentiable on the interval [a,b].

Answer 4

The second part of your question seems to be L'Hopital's rule. Done it?

Answer 5

Nooo, I haven't. Should I look up a standard proof of that?

Answer 6

http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule#Special_case

Answer 7

Basically, because f(0) = f'(0)= f''(0) = f'''(0)...... = f^(n-1)(0) = 0
g(0) = g'(0) = g''(0) = g'''(0) ........ = g^n-1(0) = 0
\[\lim_{x \rightarrow 0}(\frac{f(x)}{g(x)})=\lim_{x \rightarrow 0}(\frac{f^n(x)}{g^n(x)})\]

Answer 8

That's why it's important that \[b_n \ne 0\]

Answer 9

'Show that there exists a d>0 such that for all x with
0<mod(x)<d'

I've not done real analysis, but isn't this a bit trivial as long as x isn't equal to 0?

Answer 10

ah okay, and is \[\lim_{x \rightarrow 0}f ^{n}(x)=f^{n}(0) \] as it is continuous at 0?

Answer 11

Yes

Answer 12

And probably! I'm just having trouble showing it. I thought of maybe going from the epsilon delta definition of continuity but I didn't really get anywhere.

Answer 13

L'hopital is basically differentiating top and bottom (if the limit of both is 0) until the denominator isn't 0 (i.e. the limit itself isn't discontinuous)

Answer 14

Ah okay, that's quite cool (:

Answer 15

Is the comment that g(x) is not 0 attached to the 1st or 2nd question?

Answer 16

The first, I think it's asking me to show that g(x) is not 0 for some d, with 0<mod(x)<d

Answer 17

I misunderstood earlier.

Answer 18

Sorry, I don't know how to help with the 1st question. Intuitively it's obviously true, but I' not sure how to prove it.

Answer 19

No problem, thank you so much for the help with the other part!