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.

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.
