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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.

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marked as duplicate by Alex M., jameselmore, Johannes Kloos, Rory Daulton, Shailesh Dec 30 '15 at 0:02

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  • $\begingroup$ What is $ a_n $? What is the behaviour towards infinity? $\endgroup$ – Matt B. Sep 7 '14 at 11:31
  • $\begingroup$ $a_n$ is any constant. It has a finite value I guess. $\endgroup$ – sourish Sep 7 '14 at 11:33
  • $\begingroup$ @MattB. : $a_n$ would have any "behaviour towards infinity" since it is $x$, not $n$, that is approaching infinity. $\endgroup$ – Michael Hardy Sep 7 '14 at 12:57
  • $\begingroup$ @MichaelHardy You're right, I thought the product was infinite as well. It doesn't depend on the behaviour of the $a_i$. $\endgroup$ – Matt B. Sep 7 '14 at 13:02
  • $\begingroup$ Typo: I meant: would NOT have any "behaviour towards infinity". $\endgroup$ – Michael Hardy Sep 7 '14 at 19:16
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\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}

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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 $$ \frac{a_{1}+a_{2}+\ldots+a_{n}}{n}. $$

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$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$ $$\lim_{x\to\infty}\left[x - x\sqrt[n]{\left(1 - \frac{a_1}x\right)\left(1 - \frac{a_2}x\right)\ldots\left(1 - \frac{a_n}x\right)}\right]$$ $$\lim_{x\to\infty}\left[x - x\sqrt[n]{\left(1 - \frac{a_1+a_2+\cdots+a_n}x+O\left(\frac1{x^2}\right)\right)}\right]$$ 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 $\bf L$

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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.

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    $\begingroup$ That's incorrect. Your main term is indeed $x^n$ but it gets cancelled out by the $x$, so you have to look at the other terms as well. $\endgroup$ – Matt B. Sep 7 '14 at 12:55

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