Solving $\lim \limits _{x \to \infty} (\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x)$ $$\lim \limits _{x \to \infty}\bigg(\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x\bigg)$$
We can see the limit is of type $\infty-\infty$. I don't see anything I could do here. I can only see the geometric mean which is the $n$-th root term. Can I do anything with it? Any tips on solving this?
 A: You may write, as $x \to \infty$,
$$
\begin{align}
(x+a_1)\cdot(x+a_2)\cdot...\cdot(x+a_n)&=x^n+(a_1+a_2+\cdots+a_n)\cdot x^{n-1}+O(x^{n-2})\\\\
&=x^n\left(1+\frac{a_1+a_2+\cdots+a_n}{x}+O\left(\frac1{x^2}\right)\right)\\\\
\end{align}
$$ giving
$$
\begin{align}
\sqrt[n]{(x+a_1)\cdot(x+a_2)\cdot...\cdot(x+a_n)}&=x\left(1+\frac{a_1+a_2+\cdots+a_n}{x}+O\left(\frac1{x^2}\right)\right)^{1/n}\\\\
&=x+\frac{a_1+a_2+\cdots+a_n}{n}+O\left(\frac1{x}\right)\end{align}
$$ and

$$
\lim_{x \to \infty}\left(\sqrt[n]{(x+a_1)\cdot(x+a_2)\cdot...\cdot(x+a_n)}-x\right)=\frac{a_1+a_2+\cdots+a_n}{n}.
$$ 

A: It's more complicated to write in LaTeX, than to solve. Remember that
$$A-B = \frac {A^n - B^n} {A^{n-1} + A^{n-2} B + \dots + A B^{n-2} + B^{n-1}} .$$
Choosing $A = \sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}$ and $B = \sqrt[n] {x^n}$, note that the largest power of $x$ in $A^n - B^n$ is $x^{n-1}$, and it is multiplied by the coefficient $a_1 + \dots + a_n$.
Similarly, looking for the dominant part of $x$ in the denominator, write
$$A^{n-k} \cdot B^{k-1} = [ (x+a_1) (x+a_2) \dots (x+a_n) ] ^{\frac {n-k} n} \cdot (x^n) ^{\frac {k-1} n} = \\ (x^n) ^{\frac {n-k} n} \left[ \left( 1 + \frac {a_1} x \right) \dots \left( 1 + \frac {a_n} x \right)\right] ^{\frac {n-k} n} \cdot (x^n) ^{\frac {k-1} n} = x^{n-1} \left[ \left( 1 + \frac {a_1} x \right) \dots \left( 1 + \frac {a_n} x \right)\right] ^{\frac {n-k} n} .$$
Note that the part between square brackets tends to $1$ when $x \to \infty$ (because $n$, the number of factors, stays fixed and each factor tends to $0$), so $\dfrac {A^{n-k} \cdot B^{k-1}} {x^{n-1}} \to 1$ (i.e $A^{n-k} \cdot B^{k-1}$ behaves asymptotically like $x^{n-1}$ when $x \to \infty$). There are $n$ such terms in the denominator, so $\dfrac {n x^{n-1}} {\text{denominator}} \to 1$.
Putting everything together you get that
$$\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)} - \sqrt[n] {x^n} = A-B = \\ \frac {(a_1 + \dots a_n) x^{n-1} + \text{smaller powers of} \ x} {n x^{n-1}} \frac {n x^{n-1}} {\text{denominator}} \to \frac {a_1 + \dots + a_n} n .$$
