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Let $a_0, a_1,\ldots,a_{n-1}, a_n$ complex numbers with $a_n\neq 0$. If $$f(z)=\left|a_n+\frac{a_{n-1}}{z}+\cdots+\frac{a_{0}}{z^{n}}\right|$$ Exist $\,\,\,\,\displaystyle{\lim_{|z|\rightarrow \infty}f(z)}\,$?

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You've gotten some nice answers to questions you've asked in the past. Consider accepting them. – user17794 Oct 2 '12 at 1:46
as I do that? accepting the responses – Roiner Segura Cubero Oct 2 '12 at 1:49
@Andres When people take the effort to answer your questions, it's polite to select the response that you like and accept it as an official answer to your question. You can do that by clicking the checkmark beside the answer. – EuYu Oct 2 '12 at 1:50
up vote 2 down vote accepted

If we let $z\rightarrow \infty$, intuitively then we expect the terms containing $z$ in the denominator to drop out leaving the limit as $|a_n|$. Let's see how our intuition holds out. We have $$\left|\ \left|a_n + \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right|-|a_n|\ \right|\le\left|\frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right|$$ The above inequality follows from the reverse triangl inequality. Now apply the regular triangle inequality $$\left|\frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right|\le\left|\frac{a_{n-1}}{z}\right| + \cdots + \left|\frac{a_0}{z^n}\right|$$ If the proof is rather informal at this point you can just say that as $z\rightarrow \infty$ the right hand side above tends to $0$ and the limit holds. Otherwise, it wouldn't be too difficult to follow up with a formal $\epsilon-\delta$ proof.

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