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I need to show that $\phi_n(z)=\ln(1+z^n)$ as $z \rightarrow 0$ is an asymptotic sequence, i.e. to show that $$\lim_{z\rightarrow 0}\frac{\phi_{n+1}(z)}{\phi_n(z)}=0.$$

Is it sufficient for me to say that as $z\rightarrow 0$, $$\frac{\phi_{n+1}}{\phi_n}=\frac{\ln(1+z^{n+1})}{\ln(1+z^n)} \rightarrow 0?$$ Because $z^{n+1}$ and $z^n \rightarrow 0$ as $z\rightarrow 0$, and $\ln(1)=0?$

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up vote 2 down vote accepted

No, the $0/0$ form is indeterminate so you can't simply plug the limits into the logarithm. Consider the limits of things like $z^2/z$, $z/z$, and $z/z^2$: in each case numerator and denominator $\to0$, but the limits of them are $0$, $1$ and $\infty$ respectively. This looks like an excellect application of l'Hospital's.

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Thanks for your comments. Sorry, but how would I relate what you've said to use it in the question? Im not exactly sure how to show it converges to zero. –  Heijden Feb 23 '12 at 0:34
@Heijden: My comments simply say that pointing out num and denom go to zero does not suffice to say their ratio goes to zero. As for showing it does go to zero in this case, with valid reasoning, did you look at the link I posted? –  anon Feb 23 '12 at 0:36
Ahh okay sorry I was thinking that you where referring that I use those examples to somehow prove the limit. Yes, I did look at the link (L Hopital), but exactly what am I meant to use there? How would I apply it in my case? Do I have to differentiate it? –  Heijden Feb 23 '12 at 0:47
@Heijden: First off, you have a ratio of two functions both of which go to $0$ as $z\to0$. These are the numerator and denomenator. The rule basically says that $\lim\; f(x)/g(x) = \lim f'(x)/g'(x)$. You're trying to figure out the limit on the left side, the rule says it equals what's on the right side. Can you figure out the right side if $f(z)=\log(1+z^{n+1})$ and $g(z)=\log(1+z^n)$? –  anon Feb 23 '12 at 1:02
Thanks anon, I and shall work on it, appreciated! –  Heijden Feb 23 '12 at 18:34
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Just use the formula $\log (1+w) = w + \mathcal{O}(w^2)$,as $\mathbb C \ni w \to 0$.

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