Compute $\lim_{n\to1}(n^5-(n^3-1)^{-1})$

Question

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

Answer 1

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

Answer 2

$$\lim_{ n\to 1}\frac{n^5-1}{n^3-1}=\lim_{ n\to 1}\frac{(n-1)(n^4+n^3+n^2+n+1)}{(n-1)(n^2+n+1)}=\lim_{ n\to 1 }\frac{(n^4+n^3+n^2+n+1)}{(n^2+n+1)}$$ 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 $x=1$.

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For the modified case,

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

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

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