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..
 A: If $x \neq 1$, we have ${x^n-1 \over x -1} = 1+x+ \cdots + x^{n-1}$.
A: division of polynomials or finite induction
$$
x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1)
$$
Thus
$$
\lim_{x \to 1}\dfrac{x^n - 1}{x - 1} = \lim_{x \to 1} \dfrac{(x-1)(x^{n-1} + x^{n-2} + \ldots + x^2 + x + 1)}{x - 1} = n
$$
A: Note that if $f(x)=x^n$ then
$$f'(1)=\lim_{x\to 1}\frac{x^n-1}{x-1}$$
How $$f'(x)=nx^{n-1}$$  then $$f'(1)=n\cdot(1^{n-1})=n$$
A: HINT:
suppose $y=x-1$
Then
$$\lim_{x\to 1}\frac{x^n-1}{x-1}=\lim_{y\to 0}\frac{(y+1)^n-1}{y}$$
A: Let's try an inductive proof. We first prove the result for positive integers $n$ by using induction on $n$. For $n = 1$ we can see that $$\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = \lim_{x \to 1}\frac{x - 1}{x - 1} = 1 = n$$ so that the result is verified for $n = 1$. Let's assume it is true for $n = m$ and we try to establish it for $n = m + 1$. We have $$\begin{aligned}L\,&= \lim_{x \to 1}\frac{x^{n} - 1}{x - 1}\\
&= \lim_{x \to 1}\frac{x^{m + 1} - 1}{x - 1}\\
&= \lim_{x \to 1}\frac{x^{m + 1} - x + x - 1}{x - 1}\\
&= \lim_{x \to 1}\frac{x(x^{m} - 1) + x - 1}{x - 1}\\
&= \lim_{x \to 1}x\cdot\frac{x^{m} - 1}{x - 1} + 1\\
&= 1\cdot m + 1 = m + 1 = n\end{aligned}$$ so that the result is true for $n = m + 1$ as well. By induction it is true for all positive integers $n$.
To handle negative integers we again apply induction. This time we assume that result holds for some integer $n = m$ and show that it holds for $n = m - 1$ also. We have $$\begin{aligned}L\,&= \lim_{x \to 1}\frac{x^{n} - 1}{x - 1}\\
&= \lim_{x \to 1}\frac{x^{m - 1} - 1}{x - 1}\\
&= \lim_{x \to 1}\frac{x^{m} - x}{x(x - 1)}\\
&= \lim_{x \to 1}\frac{x^{m} - 1 - (x - 1)}{x(x - 1)}\\
&= \lim_{x \to 1}\frac{1}{x}\cdot\frac{x^{m} - 1}{x - 1} - \frac{1}{x}\\
&= 1\cdot m - 1 = m - 1 = n\end{aligned}$$ So we can always go from $n = m$ to the previous integer $n = m - 1$. We know that result is true for $n = 1$ so it is true for previous integer $0$ and so true for previous integer $-1$ and so on. By induction it is true for all negative integers $n$ (as well as for $n = 0$).
Other answers given above are quite standard and well known and I have tried to present a different approach. Also all the other answers seem to have left the case when $n < 0$ although this is a simple extension once the result is established for $n > 0$.
A: Use that: $$x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)$$ The $(x-1)$ will cancel out. Can you do it now?
