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Calculate $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$

This is a problem which I found in a math textbook some time ago. I tried solving it but I did not manage to solve it as quickly as I wanted and so I lost patience. Therefore I posted it in MathSE. I liked quite a lot one of the solutions I got, so I wrote it on a sheet of paper and kept it.

Here is the full story...


Find $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$,

where $x$ is a real variable and $n$ is a fixed natural number $n \ge 1$.


$f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$

We now calculate:

$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}) \cdot (1+\frac{a_n}{x})}= \sqrt[n]{1} = 1 \tag{1}$$

We now form the below difference and we rework it a bit:

$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$

We note here that the numerator $P(x) = f(x)^n-x^n$ is a polynomial of degree $n-1$ and it has a leading coefficient of $c = a_1+a_2+\dots+a_n$. The denominator is composed of the sum of $n$ terms of the form $f(x)^{n-1-k}x^k$ (for $k=0,1, \dots, n-1$).

So we divide the numerator and the denominator by $x^{n-1}$ and we get:

\lim_{x\to \infty}f(x)-x & = \lim_{x\to \infty} \frac {\frac{f(x)^n-x^n}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]
&= \lim_{x\to \infty} \frac {\frac{P(x)}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]
&= \frac{c}{\underbrace{1 + 1 + \dots + 1}_{n\text{ times}}} \\[10pt]
&= \frac{a_1+a_2+\dots+a_n}{n}\end{align*}

The cool thing here is that the limit value happens to be the arithmetic mean of the numbers $a_1, a_2, \dots a_n$.

See also Math SE question/solution here

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