Evaluate $\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$ Evaluate the following the limit:
$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$
I tried expressing the limit in the form $f(x)g(x)\left[\frac{1}{f(x)} - \frac{1}{g(x)}\right]$ but it did not help.
 A: \begin{align}
& \lim_{x\to\infty} x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)} \\
= {} & \lim_{x\to\infty} x\left(1-\sqrt[n]{\left(1 - \frac{a_1}{x}\right)\left(1 - \frac{a_2}{x}\right)\ldots\left(1 - \frac{a_n}{x}\right)} \right) \\
= {} & \lim_{x\to\infty} x\left(1-1 +\frac{1}{n}\frac{\sum a_i}{x}\right)\\
= {} & \frac{1}{n}{\sum a_i}
\end{align}
A: Write $p(x) = (x-a_{1})(x-a_{2})\cdots (x-a_{n})$. Now
$$
\lim_{x\to\infty}[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}] = \lim_{x \to \infty}[x-\sqrt[n]{p(x)}] \\
= \lim_{x \to \infty}\left((x-\sqrt[n]{p(x)})\frac{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}\right) \\
= \lim_{x \to \infty}\left(\frac{x^{n}-p(x)}{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}\right) \\
= \lim_{x \to \infty}\left(\frac{(a_{1}+a_{2}+\ldots+a_{n})x^{n-1}+q(x)}{x^{n-1}\left(1+\frac{\sqrt[n]{p(x)}}{x}+\ldots+\frac{(\sqrt[n]{p(x)})^{n-2}}{x^{n-2}}+\frac{(\sqrt[n]{p(x)})^{n-1}}{x^{n-1}}\right)}\right) \\
= \lim_{x \to \infty}\left(\frac{a_{1}+a_{2}+\ldots+a_{n}+\frac{q(x)}{x^{n-1}}}{1+\frac{\sqrt[n]{p(x)}}{x}+\ldots+\left(\frac{\sqrt[n]{p(x)}}{x}\right)^{n-2}+\left(\frac{\sqrt[n]{p(x)}}{x}\right)^{n-1}}\right)
$$
where $q(x)$ is a polynomial of degree at most $n-2$.
Now $\frac{\sqrt[n]{p(x)}}{x} \to 1$, and $\frac{q(x)}{x^{n-1}} \to 0$ as $x \to \infty$ so the whole expression tends to
$$
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}.
$$
A: $$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$
$$\lim_{x\to\infty}\left[x - x\sqrt[n]{\left(1 - \frac{a_1}x\right)\left(1 - \frac{a_2}x\right)\ldots\left(1 - \frac{a_n}x\right)}\right]$$
$$\lim_{x\to\infty}\left[x - x\sqrt[n]{\left(1 - \frac{a_1+a_2+\cdots+a_n}x+O\left(\frac1{x^2}\right)\right)}\right]$$
Using taylor's theorem:
$$(1+x)^p=1+(p-1)x+\frac{p(p-1)}2x^2+\cdots$$
You end up with:
$$\lim_{x\to\infty}\left[x\left( \frac1n\frac{a_1+a_2+\cdots+a_n}{x}+O\left(\frac1{x^{2}}\right)\right)\right]$$
Which surely is:
$$nL=a_1+a_2+\cdots+a_n$$
where the limit is $\bf L$
A: You know that the product $(x - a_1)(a - a_2)\cdots(x - a_n)$ is equal to $x^n + \cdots + a_1a_2\cdots a_n$, where the $\cdots$ between $x^n$ and the product of the $a_i$'s is made of powers of $x$ with an exponent that is smaller than $n$. After you've seen this, it's easy to see that the only term that matters inside that root is $x^n$, because all the other ones can be left out when $x$ goes to $\infty$.
So your limit is:
$$\lim_{x\to\infty} (x - \sqrt[n]{x^n}),$$
and now it will all depend on whether $n$ is even or odd. If it's even, then $\sqrt[n]{x^n} = \lvert x\rvert$, otherwise $\sqrt[n]{x^n} = x$. So, if $x$ goes to $-\infty$, for example:
$$\lim_{x\to-\infty} (x - \sqrt[n]{x^n}) = \begin{equation}
   \begin{cases}
   -\infty, & \text{if $n$ is even;} \\0, & \text{if $n$ is odd.}
   \end{cases}
\end{equation}$$
Otherwise, if $x$ goes to $+\infty$, it's easy to see that the limit is $0$ in both cases.
