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**sfspitfire23** Prove: $\displaystyle \frac{n^2+1}{n-2}$ diverges

For every L there exists an epsilon greater than zero, for every N greater than 0, n>0.

So, $\displaystyle |\frac{n^2+1}{n-2}-L|>\epsilon$

Pf: Let $\displaystyle L$ be given. Set $\displaystyle \epsilon=.001$ and let $\displaystyle N>0$ be given. There exists $\displaystyle n_0>N$ s.t. $\displaystyle n_0=3$. Then, if $\displaystyle |\frac{n_0^2+1}{n_0-2}-L|<\epsilon$, we would have $\displaystyle |10-L|<\epsilon$ $\displaystyle \implies$ $\displaystyle -.001<L-10<.001 \implies 9.999<L<10.001$. But then if $\displaystyle |\frac{n_{0+1}^2+1}{n_{0+1}-2}-L|<\epsilon$ we would have |8.5-L|<\epsilon. So, $\displaystyle 8.499<L<8.501$. But then $\displaystyle L$ would be in different intervals. Thus the sequence diverges as desired.

Is this the correct way to go about this?

Thx!