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Compute $\displaystyle\lim_{n \to 1}\left(n^5 - \cfrac1{n^3 - 1}\right)$. It's undefined, but how do I show it?

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$$\lim_{ n\to 1}\frac{n^5-1}{n^3-1}=\lim_{ n\to 1}\frac{(n-1)(n^4+n^3+n^2+n+1)}{(n-1)(n^2+n+1)}=\lim_{ n\to 1 }\frac{(n^4+n^3+n^2+n+1)}{(n^2+n+1)}$$ as $n\to1\implies n\ne 1,n-1\ne 0$

$$\lim_{ n\to 1}\frac{n^5-1}{n^3-1}=\frac5 3$$, so the limit does exist at $x=1$.

For the modified case,

$\lim_{n \to 1}\left(n^5 - \cfrac1{n^3 - 1}\right)=\lim_{n \to 1}\left(\cfrac{n^8-n^5 - 1}{n^3-1}\right)$

$\lim_{n \to 1^+}\left(n^5 - \cfrac1{n^3 - 1}\right)=-\infty$

$\lim_{n \to 1^-}\left(n^5 - \cfrac1{n^3 - 1}\right)=\infty$

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For $n\to 1$, yes, but $n\to\infty$ was the question. – Berci Oct 21 '12 at 9:45
@Berci, look at the original question that was edited by Hagen – lab bhattacharjee Oct 21 '12 at 9:47

Hint: $\frac{n^5-1}{n^3-1}=n^2\cdot\frac{1-n^{-5}}{1-n^{-3}}$.

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N tends to 1 pls.(not infinity). – Nero Oct 21 '12 at 9:41
Then what's with $\displaystyle\lim_{n\to\infty}$ in OP's question? – Frenzy Li Oct 21 '12 at 9:42

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