Limit $\lim\limits_{x \to +\infty} \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ =? What is the limit of $ f(x) = \sqrt[n]{(x+a_1)(x+a_2)...(x+a_n)} - x$ when $x$ goes to plus infinity? The number $n$ is fixed and $a_i$ are some constants. 
All I know is the answer:
$(a_1 + a_2 + ... + a_n) / n$
I figured out that I can limit the root from above with the arithmetic mean. And this gives me some nice expression as an upper limit. But OK... this only proves the limit is $\leq (a_1 + a_2 + ... + a_n) / n$.
But how do I limit it from below with the same value. I tried using the harmonic mean but it seems it does not lead me to anything nice. So... any ideas?  

Notice that $$(x+a_1)(x+a_2)\cdot (x+a_n)=x^n+ b x^{n-1}+\mathcal{O}(x^{n-2})$$
Where $b=a_1+a_2+\dots +a_n$, and 
$$\sqrt[n]{x^n+bx^{n-1}+\mathcal{O}(x^{n-2})}=\left(x^n+bx^{n-1}+\mathcal{O}(x^{n-2})\right)^{1/n}=x \left(1+bx^{-1} +\mathcal{O}(x^{-2})\right)^{1/n}$$
Now, we can use the first order Taylor approximation of $(1+x)^q$ around $x=0$ (which is $1+q x+\mathcal{O}(x^2)$) to get 
$$x \left(1+b +\mathcal{O}(x^{2-n})\right)^{1/n}= x\left(1+\frac{b}{n x} +\mathcal{O}(x^{-2})\right)=x-\frac{b}{n } +\mathcal{O}(x^{-1}).  $$
From this, we can see that the limit is $\frac{b}{n}$. 
A more elementary solution: Let $f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$, note that 
$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+a_1x^{-1})(1+a_2x^{-1})\cdot (1+a_nx^{-1})}=1\tag{1}$$ 
Note that
$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$
The numerator is a polynomial of degree $n-1$ with a leading coefficient of $a_1+a_2+\dots+a_n$, the denominator is composed of the sum of $n$ terms $f(x)^{n-1-k}x^k$. Divide both by $x^{n-1}$ to get 
\begin{align*}\lim_{x\to \infty}f(x)-x&=\lim_{x\to \infty} \frac{f(x)^n-x^n}{x^{n-1}} \cdot \frac{1}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1}\\
&=(a_1+a_2+\dots+a_n)\cdot \frac{1}{1+1+\dots+1} \\ &
=\frac{a_1+a_2+\dots+a_n}{n}\end{align*}
