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).
 A: Without using l'Hopital (directly), but if you know derivatives, then by definition:
$$
\lim_{x \to 1} \frac{x^5-1}{x-1} = \big(x^5 - 1\big)'\bigg|_{x=1}=5x^4\bigg|_{x=1} = 5 \\[5px]
\lim_{x \to 1} \frac{x^2-1}{x-1} = \big(x^2 - 1\big)'\bigg|_{x=1}=2 x\bigg|_{x=1} = 2
$$
It follows that the given limit is:
$$
\lim_{x \to 1} \cfrac{x^5-1}{x^2-1} = \lim_{x \to 1} \cfrac{\;\;\cfrac{x^5-1}{x-1}\;\; }{ \cfrac{x^2-1}{x-1} } = \cfrac{\;\;\lim_{x \to 1} \cfrac{x^5-1}{x-1}\;\; }{ \lim_{x \to 1} \cfrac{x^2-1}{x-1} } = \frac{5}{2}
$$
A: $$\frac{x^5-1}{x^2-1}=\frac{(x-1)(1+x+x^2+x^3+x^4)}{(x-1)(x+1)}$$
and so the limit is $5/2$.
A: \begin{align}\lim_{x\to1}\frac{x^5-1}{x^2-1}&=\lim_{x\to1}\frac{(x-1)(x^4+x^3+x^2+x+1)}{(x-1)(x+1)}\\&=\lim_{x\to1}\frac{(x^4+x^3+x^2+x+1)}{(x+1)}\\&=\frac52\end{align}
A: $$\frac{x^5-1}{x^2-1}=\frac{(x-1)(x^4+x^3+x^2+x+1)}{(x-1)(x+1)}=\frac{(x^4+x^3+x^2+x+1)}{(x+1)}\to\frac52$$
A: using Horner's method:
$$x^5 -1 =(x-1)(x^4 +x^3 +x^2 +x+1)$$
$$x^2 -1=(x-1)(x+1)$$
A: 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
$$
\lim_{x\to 1}\frac{x^5-1}{x^2-1}=\lim_{x\to 1}
\frac{1}{x+1}\times
\lim_{x\to 1}\frac{x^5-1}{x-1}=\frac{1}{2}\times5=\frac{5}{2}.
$$
