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Yes, I read https://math.stackexchange.com/a/3096211/578535

Then it made me think about the generalization of the following.

$$ \lim_{x\rightarrow\infty}\left(\sqrt[n]{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}-\sqrt[n]{a_n}x\right) $$

From my problem solving experience, I could find a pattern and predict as following.

$$ \lim_{x\rightarrow\infty}\left(\sqrt[n]{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}-\sqrt[n]{a_n}x\right)=\frac{\sqrt[n]{a_n}a_{n-1}}{na_n} $$

I tried to prove using squeeze theorem but failed to find suitable expression

$$ ?<\sqrt{2x^2+4x-8}-\sqrt{2}x<\sqrt{2(x+1)^2}-\sqrt{2}x=\sqrt{2} $$

Here I did a simple one where $ n=2 $, but I think you can see what I am trying to do.

I am trying to show that only two highest terms inside the nth root matter.

Is this prediction correct? How can we prove it if it is correct?

Apparently the link I put proves this due to the Vieta's formula, but I still want different proof because I don't understand that proof.

Apparently it doesn't because in that link the polynomial can be factorised with no imaginary component.

I believe there is a different proof for this.

Pizzaroot
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2 Answers2

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$$L=\lim_{x\rightarrow \infty} \sqrt[n]{a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0}- a_n^{1/n} x.$$

$$\Rightarrow L=\lim_{x \rightarrow} a_n^{1/n} x \sqrt[n]{1+\frac{a_{n-1}}{a_n}\frac{1}{x}+\frac{a_{n-2}}{a_n}\frac{1}{x^2}+.....+\frac{a_0}{a_n}\frac{1}{x^n}}-a_n ^{1/n} x.$$ $$\Rightarrow L= \lim_{x\rightarrow \infty} a_n^{1/n}x \left[1+\frac{1}{n}\frac{a_{n-1}}{a_n}\frac{1}{x}+.....+\frac{1}{n}\frac{a_{0}}{a_n} \frac{1}{x^n} \right]-a_n^{1/n}x$$ In above we have used binomial approximation that $(1+z)^k\approx 1+kz$, if $|z|<<1$. So we get $$L=\lim_{x\rightarrow \infty} \left(a_n^{1/n} x- \frac{1}{n} a_n^{1/n}\frac{a_{n-1}}{a_n}+(.)\frac{1}{x}+---+(.)\frac{1}{x^{n-1}}-a_n^{1/n} x \right)=\frac{1}{n}~a_n^{(1-n)/n}~a_{n-1}.$$

Z Ahmed
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Here is a low-tech solution: I will assume that $a_n > 0$. Define $q(t)$ by

$$ q(t) = \frac{a_n + a_{n-1}t + \cdots + a_{1}t^{n-1} + a_{0}t^n}{a_n}. $$

Then $q(0) = 1$ and $q'(0) = a_{n-1}/a_n$. Also, we can write

$$ a_n x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_0 = a_n x^n q\left(\frac{1}{x}\right). $$

Plugging this to the limit in question, we get

\begin{align*} &\lim_{x\to\infty} \left( \sqrt[n]{a_n x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x + a_0} - \sqrt[n]{a_n x^n} \right) \\ &\quad = \sqrt[n]{a_n} \lim_{x\to\infty} x \left( \sqrt[n]{\smash[b]{q(1/x)}} - \sqrt[n]{\smash[b]{q(1)}} \right) \\ &\quad = \sqrt[n]{a_n} \lim_{t\to0^+} \frac{\sqrt[n]{\smash[b]{q(t)}} - \sqrt[n]{\smash[b]{q(1)}}}{t} \tag{$t=1/x$} \\ &\quad = \sqrt[n]{a_n} \, \left. \frac{d}{dt} \sqrt[n]{\smash[b]{q(t)}} \right|_{t=0} \\ &\quad = \sqrt[n]{a_n} \, \left. \frac{q'(t)}{nq(t)^{(n-1)/n}}\right|_{t=0} \\ &\quad = \frac{\sqrt[n]{a_n} \, a_{n-1}}{n a_n} \end{align*}

Of course, if the Big-Oh notation is available, we have the following shorter proof:

\begin{align*} \sqrt[n]{a_n x^n + \cdots} - \sqrt[n]{a_n x^n} &= \sqrt[n]{a_n} \, x \left( \sqrt[n]{1 + \frac{a_{n-1}}{a_n x} + \mathcal{O}\left(\frac{1}{x^2}\right)} - 1 \right) \\ &= \sqrt[n]{a_n} \, x \left( \left(1 + \frac{a_{n-1}}{na_n x} + \mathcal{O}\left(\frac{1}{x^2}\right) \right) - 1 \right) \\ &= \frac{\sqrt[n]{a_n} \, a_{n-1}}{n a_n} + \mathcal{O}\left(\frac{1}{x}\right), \end{align*}

which converges to the desired limit as $x\to\infty$.

Sangchul Lee
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