How to find the limit of $|f(z)|=|z^{20}+z+1|$ as $|z|\to \infty$ 
Find the limit of $\lim_{|z|\to \infty}|f(z)|$ where$f(z)=z^{20}+z+1$.  

Well I haven't done this kind of limit problem so I have no idea how to proceed. But I came across an mathstack answer (I can't find it anymore), where a same kind of problem was solved by taking the transformation $w=1/z$. Since $|z|\to \infty$ so $w\to 0$. Following that process, in my problem- 
$$\lim_{|z|\to \infty}|f(z)|=\lim_{w\to 0}|f(1/w)|=\lim_{w\to 0}|\frac{1}{w^{20}}+\frac{1}{w}+1|$$ which gives the limit as $\infty.$ 
Is it correct? I didn't understand the method. Can somebody give me the concept behind this, if this method is correct? If not, then what other method I can use? Thanks.
 A: $|f(z)|=|z|^{20}|(1+\frac 1 {z^{19}}+\frac 1 {z^{20}})| \to \infty$ because $1+\frac 1 {z^{19}}+\frac 1 {z^{20}} \to 1$ as $|z| \to \infty$.
This works for any non-constant polynomial!. 
A: With
$f(z) = z^{20} + z + 1, \tag 1$
we may write, for $z \ne 0$,
$f(z) = z^{20}(1 + z^{-19} + z^{-20}), \tag 2$
whence
$\vert f(z) \vert = \vert z^{20} \vert \vert 1 + z^{-19} + z^{-20} \vert = \vert z \vert^{20} \vert 1 + z^{-19} + z^{-20} \vert; \tag 3$
also,
$\vert  1 + z^{-19} + z^{-20} \vert = \vert 1 - (-z^{-19} -z^{-20}) \vert$
$\ge \vert \vert 1 \vert - \vert -z^{-19} - z^{-20} \vert \vert = \vert 1 - \vert z^{-19} + z^{-20} \vert \vert; \tag{4}$
now for any $0 < \epsilon < 1/2$, for $\vert z \vert$ sufficiently large, we have both
$\vert z \vert^{-19}, \vert z \vert^{-20} < \epsilon; \tag 5$
thus,
$\vert z^{-19} + z^{-20} \vert \le \vert z \vert^{-19} + \vert z \vert^{-20} < 2\epsilon < 1, \tag 6$
and we infer that
$\vert 1 - \vert z^{-19} + z^{-20} \vert \vert > 1 - 2\epsilon > 0; \tag 7$
combining (3), (4) and (7) we see that
$\vert f(z) \vert > \vert z \vert^{20} (1 - 2\epsilon) > 0; \tag 8$
it now follows readily that
$\displaystyle \lim_{\vert z \vert \to \infty } \vert f(z) \vert = \infty. \tag 9$
$OE\Delta$.
A: The next step can be to write 
$$\lim_{w\to0}\left|\frac{1}{w^{20}}+\frac{1}{w}+1\right|=\lim_{w\to0}\frac{|1+w^{19}+w^{20}|}{|w|^{20}}$$
and compute instead the limit $\lim_{w\to0}\frac{|w|^{20}}{|1+w^{19}+w^{20}|}=0$
Since $\lim_{w\to0}w^{20}=0$ and $\lim_{w\to0}\left|1+w^{19}+w^{20}\right|=1$, it follows that the limit of their quotient is $0/1=0$.
Therefore, the limit of the reciprocal of their quotient is $\infty$.
