# Showing something is an asymptotic sequence

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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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.
@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
Just use the formula $\log (1+w) = w + \mathcal{O}(w^2)$,as $\mathbb C \ni w \to 0$.