Limit of $\frac{x^5-1}{x^2-1}$ I need to determine if the following limit exists.
$$\lim_{x\to 1}\frac{x^5-1}{x^2-1}$$
I've already proved using L'Hospital that this limit exists and should equal to $\frac{5}{2}$, but unfortunately I'm not allowed to used anything more than basic analysis for functions, i.e. basic definitions of convergence(at most continuity).

Note that
$$
\lim_{x\to 1}\frac{x^5-1}{x-1}=f'(1)=5
$$
where $f(x)=x^5$ (so $f'(x)=5x^4$) by the definition of the derivative. Hence
$$
\lim_{x\to 1}\frac{x^5-1}{x^2-1}=\lim_{x\to 1}
\frac{1}{x+1}\times
\lim_{x\to 1}\frac{x^5-1}{x-1}=\frac{1}{2}\times5=\frac{5}{2}.
$$
