Regarding interchanging limits. $(1)\;$ We have for $f$ entire function and positive integer $m$
$\rightarrow\quad\lim_\limits{z\to \infty} \left(\dfrac{f(z)}{z^m}\right)=1$ .
I am trying to get the function $f\;$ by$f(z)= \displaystyle\sum^{\infty}_{n=0} a_{n}z^n\;$ for all $\;z\in\mathbb C$ (as it is entire).
Define $S_n(z) =\displaystyle\sum^{n-1}_{i=0} a_{i}z^i\;$ for all $\;z\in\mathbb C$.
Now consider $(S_n)_{n\in\mathbb N}$ as a sequence of function with domain set of complex number. (We are going to use uniform convergence)
Clearly $\;f(z)=\lim_\limits{n\to\infty} S_n(z)\;$ and we also have $\;(S_n)_{n\in\mathbb N}\;$ converges to $f$ uniformly on $\mathbb C$.
By dividing $z^m$ and applying limit we have
$\lim_\limits{z\to\infty}\left(\dfrac{f(z)}{z^m}\right)=\lim_\limits{z\to\infty}\left(\lim_\limits{n\to\infty}\displaystyle\frac{S_n(z)}{z^m}\right)$
Now as L.H.S is $1$, so the R.H.S is.
But to proceed further can we interchange limit in above equation? I am confused about it because we don’t know wheather $\;\lim_\limits{z\to \infty} S_n(z)\;$ exist for all $\;n\in\mathbb N\;$ or not.Without knowing it, we cannot interchange the limits and write $\;\lim_\limits{n\to\infty}\left(\lim_\limits{z\to\infty}S_n(z)\right)$.
My question is “am I right”? Or can we interchange limits? If we can’t, then how to proceed further ?
 A: This question is related to this previous question Regarding existence of limit in sequence of function.
We do know that $\lim_{z\to\infty}S_n(z)$ exists for all $n$ - the $S_n$ are all just polynomials, and will all go to $\infty$. That is not what you really want to know, though. You want to know if $\lim_{z\to\infty}\frac{S_n(z)}{z^m}$ exists for all $n$ and $m$ - and this is in fact true, (being careful not to consider the case of $z=0$ when taking the limit). It will either be $0,\infty$ or $a_m$, in the respective cases of $m>n+1$, $m<n+1$ (and some $a_k\neq 0$ for $m<k<n$) and $m=n+1$.
Thus, you can swap the limits, and the above should be a great hint for proving what Fred mentioned.
Note that the way you posed the previous question was a far too general case. What really saves the day here is the details - being an entire function, the partial sums being polynomials, and so on. Remember to include all assumptions for future questions.
A: From
$$\lim_\limits{z\to \infty} \left(\dfrac{f(z)}{z^m}\right)=1$$
we get, that there is some $r>0$ such that
$$|f(z)|/|z|^m  \ge 1/2$$
for $|z| \ge r.$ Hence
$$|f(z)| \ge 1/2 |z|^m$$
for $|z| \ge r.$
Now let $g(z):=f(1/z)$ for $z \ne 0$.
Then we get
$$|g(z)| \ge 1/2 \frac{1}{|z|^m}$$
for $|z| \ge 1/r$. Thus $g(z) \to \infty$ as $z \to 0.$ Hence $g$ has a pole in $0$. Therefore $f$ is a polynomial.
It is your turn to shoe that $f$ has the form
$$f(z)=z^m+a_{m-1}z^{m-1}+...+a_1z+a_0.$$
