$\lim_{x\rightarrow\infty}p\left(x\right)^{\frac{1}{\ln x}}$ where $p(x)$ is a polynomial of degree $n$

Question

I'd like to solve the following limit: $$\lim_{x\rightarrow\infty}p\left(x\right)^{\frac{1}{\ln x}}$$ where $p(x)$ is a polynomial of degree $n$ and we may assume $a_{n}\neq 0$

My work so far: $$\lim_{x\rightarrow\infty}p\left(x\right)^{\frac{1}{\ln x}}=\lim_{x\rightarrow\infty}\exp\left(\frac{\ln p\left(x\right)}{\ln x}\right)=\exp\left(\lim_{x\rightarrow\infty}\frac{\ln\left(p\left(x\right)\right)}{\ln x}\right)=\exp\left(\lim_{x\rightarrow\infty}\frac{\ln\left(p\left(x\right)\right)}{\ln x}\right)\overbrace{=}^{_{L'Hopital}}=\exp\left(\lim_{x\rightarrow\infty}\frac{\frac{p'\left(x\right)}{p\left(x\right)}}{\frac{1}{x}}\right)=\exp\left(\lim_{x\rightarrow\infty}\frac{xp'\left(x\right)}{p\left(x\right)}\right)=\exp\left(\lim_{x\rightarrow\infty}\frac{x\left(a_{1}+2a_{2}x+3a_{3}x^{2}\ldots+na_{n}x^{n-1}\right)}{a_{0}+a_{1}x+\ldots+a_{n}x^{n}}\right)$$

But I am not quite sure how to continue

Answer 1

Note that $$\begin{align}\lim_{x\rightarrow\infty}\frac{x\left(a_{1}+2a_{2}x+3a_{3}x^{2}+\cdots+na_{n}x^{n-1}\right)}{a_{0}+a_{1}x+a_2x^2+\cdots+a_{n}x^{n}}&=\lim_{x\to\infty}\frac{a_1x+2a_2x^2+\cdots+na_nx^n}{a_0+a_1x+a_2x^2+\cdots+a_nx^n}\\&=\lim_{x\to\infty}\frac{\frac{a_1}{x^{n-1}}+\frac{2a_2}{x^{n-2}}+\cdots+na_n}{\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\frac{a_2}{x^{n-2}}+\cdots+a_n}\\&=\lim_{x\to\infty}\frac{0+0+\cdots+na_n}{0+0+0+\cdots+a_n}=n\end{align}$$ so $$\lim_{x\rightarrow\infty}p\left(x\right)^{\frac{1}{\ln x}}=e^n$$

Answer 2

Notice that your final limit is the same as

$$\lim_{x \to \infty} \frac{na_n + (n - 1) a_{n - 1} \frac 1 x + \cdots + a_1 \frac 1 {x^{n - 1}}}{a_n + a_{n - 1} \frac 1 x + \cdots + a_0 \frac 1 {x^n}} = n.$$

In general, any rational function $R(x) = P(x) / Q(x)$ with $P, Q$ polynomials of the same degree can be handled in this manner. The limit at $\infty$ is just the quotient of the leading coefficients.

---

You could have started computing the limit in a similar manner. Notice that you can write your polynomial $p$ as

$$p(x) = a_n x^n (1 + O(1/x))$$

so that

$$p(x)^{1/\ln x} = x^{n/\ln x} \cdot \left(a_n (1 + O(1/x))\right)^{1/\ln x}$$

The second term can be squeezed between, e.g. $2^{-1/\ln x}$ and $2^{1/\ln x}$ (which both tend to $1$) while the first term is

$$x^{n/\ln x} = e^{n \ln x / \ln x} = e^n.$$

So either way the final limit is $\boxed{e^n}.$