Prove that $\lim_{x\rightarrow 1}{\frac{x^n-1}{x-1}}=n$ for all integer n without L'Hôpital Prove that 
$\lim_{x\rightarrow1}{\frac{x^n-1}{x-1}}=n$ for all integer n without L'Hôpital.
Only things that can be used are epsilon-delta, squeeze theorem and stuff like $\lim_{x\rightarrow a}{(f(x)+g(x))} = \lim_{x\rightarrow a}{f(x)} + \lim_{x\rightarrow a}{g(x)}$.
Basically I have no idea. I do not see any way to simplify the expresion as to remove the hole, and epsilon-delta does not make any progress. Also I do not see obvious functions to serve as upper and lower bounds for the squeeze theorem. I'm sorry if it seems like I did not do anything myself yet, but that's just because nothing I tried made any substantial progress..

division of polynomials or finite induction
$$
x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1)
$$
Thus
$$
\lim_{x \to 1}\dfrac{x^n - 1}{x - 1} = \lim_{x \to 1} \dfrac{(x-1)(x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1)}{x - 1} = n
$$
