TITLE: function $e^x$ as quotient of two polynomials
QUESTION [5 upvotes]: Show that the function 
$e^x$
can not be expressed as a quotient of two polynomials.
A solution could be this:
Suppose $e^x=\frac{P(x)}{Q(x)}$ then $\quad \frac{e^x}{P(x)} = \frac{1}{Q(x)}$
We know that $e^x \quad$ tends to infinity faster than any power of $x$.
So if we do
$\lim\limits_{x\to \infty}\frac{e^x}{P(x)} = \infty $ and
$\lim\limits_{x\to \infty}\frac{1}{Q(x)} = 0 $
we get a contradiction.
Do you think is it a good solution?
and .... could anyone suggest a better solution?

REPLY [5 votes]: If you want a more formal reasoning, you can use the derivative. Let $n,m$ be the degrees of the polynomials $P_n(x),Q_m(x)$, respectively:
$$e^x=\frac{P_n(x)}{Q_m(x)}$$
Then, $e^x\,Q(x)=P(x)$ and you can differentiate both sides to obtain:
$$e^x\,Q_m(x)+e^x\,Q'_{m-1}(x)=P'_{n-1}(x)$$
so in the left-hand side you have a new polynomial of the same degree as $Q_m$, $R_m(x)=Q_m(x)+Q'_{m-1}(x)$ and:
$$e^x\,R_m(x)=P'_{n-1}(x)$$
Observe that the degree of the polynomial in the right-hand side decreases.
Repeating this process $n$ times, you finally obtain a polynomial $S_m(x)$ such that:
$$e^x\,S_m(x)=0$$
which is a contradiction.