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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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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