strange... limit

**disclaimer** · March 8th 2008, 01:01 AM

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?

Help appreciated.

**wingless** · March 8th 2008, 01:41 AM

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

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.

**mr fantastic** · March 8th 2008, 01:41 AM

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?

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

$\lim_{n\rightarrow\infty} (\sqrt{n^{10}-2n^2+2} - \sqrt{n^{10} + \text{polynomial stuff of degree less than 10}}\, )$ .

**disclaimer** · March 8th 2008, 01:50 AM

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.

**Plato** · March 8th 2008, 04:01 AM

Could it be $u_n = \sqrt[n]{{n^{10} - 2n^2 + 2}}$ ?

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