# strange... limit

• March 8th 2008, 01:01 AM
disclaimer
strange... limit
Hello;

$\lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}$

To me, it's $\infty$, as I think we can write:

$\lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}=$ $\lim_{n\rightarrow\infty}\sqrt{n^{10}\left(1-\frac{2}{n^8}+\frac{2}{n^{10}}\right)}=$ $\lim_{n\rightarrow\infty}\left(n^5\sqrt{1-\frac{2}{n^8}+\frac{2}{n^{10}}}\right)=$ $\infty\cdot1=\infty$

But the answer in my textbook is $1$. Am I missing something here? :confused:

Help appreciated. :)
• March 8th 2008, 01:41 AM
wingless
$\lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2} = \infty$

Are you sure that it's $\lim_{n\rightarrow\infty}$ instead of $\lim_{n\rightarrow 1}$ or something? Because the limit approaching to infinity is infinity (as you found), not 1.
• March 8th 2008, 01:41 AM
mr fantastic
Quote:

Originally Posted by disclaimer
Hello;

$\lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}$

To me, it's $\infty$, as I think we can write:

$\lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}=$ $\lim_{n\rightarrow\infty}\sqrt{n^{10}\left(1-\frac{2}{n^8}+\frac{2}{n^{10}}\right)}=$ $\lim_{n\rightarrow\infty}\left(n^5\sqrt{1-\frac{2}{n^8}+\frac{2}{n^{10}}}\right)=$ $\infty\cdot1=\infty$

But the answer in my textbook is $1$. Am I missing something here? :confused: Mr F says: No. But your book probably is ..... the print for the continuation of the expression that you're taking the limit of.

Help appreciated. :)

The question was probably meant to be something like

$\lim_{n\rightarrow\infty} (\sqrt{n^{10}-2n^2+2} - \sqrt{n^{10} + \text{polynomial stuff of degree less than 10}}\, )$.
• March 8th 2008, 01:50 AM
disclaimer
Well, in the book they give just $u_n=\sqrt{n^{10}-2n^2+2}$ and the problem is to find the limit of the sequence having such a general formula.

Thanks for taking the time to look into that. :)
• March 8th 2008, 04:01 AM
Plato
Could it be $u_n = \sqrt[n]{{n^{10} - 2n^2 + 2}}$?