# Possible values of $\lim_{x \rightarrow \infty} \left(\frac{p(x)}{q(x)}\right)^x$

I was asked by some freshmen the follwing two questions regarding the limit:

Using the fact that $$\lim_{x \rightarrow \infty} \left(1 + \frac{1}{x}\right)^x = e$$

evaluate $$\lim_{x \rightarrow \infty} \left(1 + \frac{3}{x^2}\right)^x$$ $$\lim_{x \rightarrow \infty} \left(\frac{x^2-2x-3}{x^2-3x-28}\right)^x$$

I got the answers $1$ and $e$ respectively. Hope nothing wrong.

In the second question, my attempt is like this:

$$\frac{x^2-2x-3}{x^2-3x-28} = \frac{(x-3)(x+1)}{(x-7)(x+4)}$$ $$\lim_{x \rightarrow \infty} \left(\frac{x-3}{x-7}\right)^x = e^{4}$$ $$\lim_{x \rightarrow \infty} \left(\frac{x+1}{x+4}\right)^x = e^{-3}$$ $$\lim_{x \rightarrow \infty} \left(\frac{x^2-2x-3}{x^2-3x-28}\right)^x = e^4e^{-3} = e$$

We can also do the other way round: $$\lim_{x \rightarrow \infty} \left(\frac{x-3}{x+4}\right)^x = e^{-7}$$ $$\lim_{x \rightarrow \infty} \left(\frac{x+1}{x-7}\right)^x = e^{8}$$

Indeed, the power of $e$ in the final answer is the difference of the sums of roots of $x^2-2x-3$ in the numerator and $x^2-3x-28$ in the denominator.

From this question, I suspect that

$$\lim_{x \rightarrow \infty} \left(\frac{x^n + \alpha_{n-1} x^{n-1} + \ldots + \alpha_0}{x^n + \beta_{n-1} x^{n-1} + \ldots + \beta_0}\right)^x = e^{\alpha_{n-1} - \beta_{n-1}}$$

is true, but I have no idea how to prove it, if the polynomials are not able to be factorized.

The freshmen had not learnt the rule yet. It would be good if there is a proof without using the rule to let them know an easy way to check their answers and why this way holds.

• Why do you doubt it? Why do you suspect it? – Thomas Andrews Dec 8 '14 at 4:10
• I added what I observed when I evaluate the limits. – Empiricist Dec 8 '14 at 4:20
• But why do you say "I doubt..." Perhaps you mean suspect? Doubt means you think the result is likely untrue... – Thomas Andrews Dec 8 '14 at 4:21
• Ooops. Sorry for misleading. I misused the word. I should say suspect. – Empiricist Dec 8 '14 at 4:22
• I tried to factorise $x^2-3x-8$ nicely but I couldn't. You mean $x^2-3x-28$? – user1537366 Jan 4 '15 at 12:49

First prove that $$\lim_{x\to\infty} \left(1+\frac{\alpha_{n-1}}{x}+\dots + \frac{\alpha_0}{x^n}\right)^x=e^{\alpha_{n-1}}$$

Then your guess follows by dividing numerator and denominator by $x^n$

More generally, you can show that if $f(0)=1$ and $f'(0)=\alpha$ with $f'(x)$ continuous at $x=0$, then $$\lim_{x\to\infty} f\left(\frac 1x\right)^x = e^\alpha$$

To prove this, note that:

$$\lim_{x\to\infty} f\left(\frac 1x\right)^x = \lim_{y\to 0^+} f(y)^{1/y}$$ and use L'Hopital to compute the limit of the log of $f(y)^{1/y}$, which is $\frac{\log f(y)}{y}$.

Technically, you don't need L'Hopital, since $$\lim_{y\to 0} \frac{\log f(y)}{y}$$ is just the definition of the derivative of $\log f(y)$ at $y=0$. You do need the derivative of the natural logarithm, though.

• Do you think the first statement can be proved without using L'Hopital rule? In fact these questions are from the freshmen, and they are not supposed to know L'Hopital rule. I just raise this to see if there is an easy way to check the answers, and I think it would be good if I can provide a readable proof for them. – Empiricist Dec 8 '14 at 4:32
• Well, $\lim_{y\to 0} \frac{\log f(y)}{y}$ is just the derivative of $\log f(y)$ at $y=0$, by definition, so L'Hopital isn't really needed. – Thomas Andrews Dec 8 '14 at 4:35
• @SRX Even if you do not have the $\log$ function yet, it is still possible to do this; see my answer. – user1537366 Jan 4 '15 at 13:30

Let $c=a_{n-1}-b_{n-1}$. Assume $c\ne 0$.

Let $p(x)=(a_{n-1}-b_{n-1})^{-1}\left((a_{n-1}-b_{n-1}) x^{n-1} + \dots + a_0-b_0\right)$.

Let $q(x)=x^n + b_{n-1} x^{n-1} + \dots + b_0$.

Let $f(x)=\left(1+c\frac{p(x)}{q(x)}\right)^\frac{q(x)}{p(x)}$.

Let $g(x)=\frac{xp(x)-q(x)}{q(x)}$.

$$\left(\frac{x^n + a_{n-1} x^{n-1} + \dots + a_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x = \left(1+\frac{(a_{n-1}-b_{n-1}) x^{n-1} + \dots + a_0-b_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x = \left(1+c\frac{p(x)}{q(x)}\right)^x=\left(\left(1+c\frac{p(x)}{q(x)}\right)^\frac{q(x)}{p(x)}\right)^{1+\frac{xp(x)-q(x)}{q(x)}}=f(x)f(x)^{g(x)}$$

$f(x)\to e^c$ as $x\to\infty$, since $\frac{q(x)}{p(x)}\to\infty$.

For large $x$, $f(x)\in e^c\times [\frac{1}{2},2]$.

For large $x$, $g(x)\in M\times[-\frac{1}{x},\frac{1}{x}]$, for some $M$.

So, for large $x$, $f(x)^{g(x)}\in [2^{-M\frac{1}{x}},2^{M\frac{1}{x}}]$, both of which $\to 1$.

So $f(x)^{g(x)}\to 1$ as $x\to\infty$ by squeeze theorem.

Hence $\left(\frac{x^n + a_{n-1} x^{n-1} + \dots + a_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x\to e^c$ as $x\to\infty$.