# Why $\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$ for polynomials $P(x)$ and $Q(x)$?

I am trying to reason why for any $$2$$ polynomials $$P(x)$$ and $$Q(x)$$ defined over the reals, $$\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$$. This assertion was made in this answer which I am presently trying to understand. Although I see that with repeated applications of differentiation to the numerator, we will eventually get a constant, the main hurdle in my seeing this is that after we have differentiated the denominator to get $$e^x \cdot Q'(e^x)$$, after applying the product rule, the summands may be of opposite sign due elements of the sequence $$Q(x), Q'(x), Q''(x),\ldots$$ varying as to whether they get positive infinite or negative infinite; consequently, the denominator may not get arbitrarily large in either a positive or negative direction, as one of the summands from applying the product rule which get arbitrarily large positive can cancel out another summand which gets arbitrarily large negative.

So, what is the reason why $$\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$$? If my hypothetical about the summands whose presence derives from applying the product rule growing in opposite directions never holds, why would that be?

• You need to add the hypothesis of $Q(x)$ to be not constant. Commented Jul 12 at 0:07

If $$Q(x)=a_0+a_1x+\cdots+a_nx^n$$, with $$n\in\{1,2,3,\ldots\}$$ and $$a_n\ne0$$, then\begin{align}\lim_{n\to\infty}\frac{Q(e^x)}{e^{nx}}&=\lim_{n\to\infty}a_0e^{-nx}+a_1e^{-(n-1)x}+\cdots+a_{n-1}e^{-x}+a_n\\&=a_n,\end{align}and therefore$$\lim_{x\to\infty}\frac{e^{nx}}{Q(e^x)}=\frac1{a_n}.$$So, if $$k\in\{0,1,2,3,\ldots\}$$,$$\lim_{x\to\infty}\frac{x^k}{Q(e^x)}=\lim_{n\to\infty}\frac{x^k}{e^{nx}}\cdot\frac{e^{nx}}{Q(e^x)}=0\cdot\frac1{a_n}=0.$$And therefore, if $$P(x)$$ is a polynomial function,$$\lim_{x\to\infty}\frac{P(x)}{Q(e^x)}=0.$$

• What if $Q$ is constant? Commented Jul 12 at 16:42
• @jjagmath I assumed that its degree is a natural number. Commented Jul 12 at 18:08
• I know from many of your posts in this forum, that you're far from been a newbie in Math. You know that the notation $\Bbb N$ is not standard, since many consider $0 \in \Bbb N$ while many others don't. So why would you do that? Commented Jul 12 at 18:46
• @jjagmath For me, $\Bbb N={1,2,3,\ldots\}$, whereas $\Bbb Z_+=\{0,1,2,3,\ldots\}$. But I've edited my answer. Commented Jul 12 at 23:09

I'm not sure I understand your objection. In general, if the numerator of a fraction grows sufficiently slower than its denominator, regardless of which infinity that denominator goes to ($$+\infty$$ or $$-\infty$$), the fraction overall goes to zero. For instance, $$\lim_{x \to \infty} \frac{\sin x}{e^x} = \lim_{x \to \infty} \frac{\sin x}{1-x} = 0$$

This is mostly just the same spirit as the solution as given by José Carlos Santos, but hopefully a tad more accessible if convoluted.

Perhaps it would be best to write out the polynomials explicitly, say $$P(x) = \sum_{n=0}^N p_n x^n \qquad Q(x) = \sum_{n=0}^M q_n x^n$$ (To avoid trivialities, also let $$M \ge 1$$, so we're not dealing with a constant polynomial, and will assume $$p_N \ne 0$$ and $$q_M \ne 0$$.) Then $$\lim_{x \to \infty} \frac{P(x)}{Q(e^x)} = \lim_{x \to \infty} \frac{\sum_{n=0}^N p_n x^n}{\sum_{n=0}^M q_n e^{nx}}$$ Clearly, $$\lim_{x \to \infty} P(x)= \operatorname{sign}(p_N) \cdot \infty$$; ultimately, it won't matter which. Meanwhile, $$e^{nx}$$ is positive for every $$n$$ and every $$x$$, and hence $$\lim_{x \to \infty} Q(x) =\operatorname{sign}(q_M) \cdot \infty$$. We can apply L'Hopital's rule, then, since we have an "$$\infty/\infty$$" type of form. (The signs of the infinities do not matter here.)

In fact, it would not be hard to justify doing it $$N$$ times, since in every case until that point we will get the same situation. This will kill all but the leading coefficient on top, i.e. it kills the prior terms of degree $$N-1$$ or less. $$\lim_{x \to \infty} \frac{P(x)}{Q(e^x)} = \lim_{x \to \infty} \frac{n! p_n}{\sum_{n=1}^M n^N q_n e^{nx}}$$

Notice that the exponential in the denominator sticks around, so there is no need to consider what happens when we differentiate some other number of times: repeated differentiation will cause polynomials to vanish, but exponentials will not. We only lose the $$n=0$$ term because that is implicitly a constant anyways.

So ultimately we get a constant on top, and a function that grows without bound on bottom. So we indeed have that the limit goes to zero.

• What if the $q_n$ are of different signs in $\sum_{n=1}^M n^N q_n e^{nx}$, such that this sum becomes $0$ or a constant in as $n$ goes to $\infty$? (I mean that even though the $e^{nx}$ gets arbitrarily large, can the signs they have from $q_n$ make some of them cancel out, like how $-5 \cdot e^{nx} + 5 \cdot e^{(n+1)x} = 0$)? Is this impossible? Commented Jul 12 at 0:58
• @PrincessMia $-5 \cdot e^{nx} + 5 \cdot e^{(n+1)x} = 0$ if and only if $x=0$.
– Gary
Commented Jul 12 at 1:38

We will use two properties.

• if $$\deg R<\deg S$$ for two polynomials $$R(x), S(x),$$ then $${R(x)\over S(x)}\underset{x\to\infty}{\longrightarrow} 0.$$
• $$e^t>t>0$$ which implies $$e^x=(e^{x/m})^m> m^{-m}x^m$$

Assume $$\deg P=m-1$$ and $$\deg Q\ge 2.$$ Denote $$y=e^x.$$ Then $${P(x)\over Q(e^x)}= {P(x)\over x^m}{y\over Q(y)}\,{x^m\over e^{x}}\quad (*)$$ The first two ratios tend to $$0.$$ Next the second property gives for $$x>0$$ $${x^m\over e^{nx}}\le {x^m\over e^x}\le m^m$$ i.e. the last ratio is bounded. Thus the limit in OP is equal $$0.$$ The remaining case $$\deg Q=1$$ follows from $$(*)$$ and the boundedness of the ratio $$y/Q(y)$$ for large $$y.$$